Copyright © 2017 Pearson Education, Inc. 170
Chapter 2 GRAPHS AND FUNCTIONS
Section 2.1 Rectangular Coordinates and Graphs
1. The point (–1, 3) lies in quadrant II in the rectangular coordinate system.
2. The point (4, 6) lies on the graph of the equation y = 3x – 6. Find the y-value by letting x = 4 and solving for y.
3 4 6 12 6 6y
3. Any point that lies on the x-axis has y-coordinate equal to 0.
4. The y-intercept of the graph of y = –2x + 6 is (0, 6).
5. The x-intercept of the graph of 2x + 5y = 10 is (5, 0). Find the x-intercept by letting y = 0 and solving for x.
2 5 0 10 2 10 5x x x
6. The distance from the origin to the point (–3, 4) is 5. Using the distance formula, we have
2 2
2 2
( , ) ( 3 0) (4 0)
3 4 9 16 25 5
d P Q
7. True
8. True
9. False. The midpoint of the segment joining (0, 0) and (4, 4) is
4 0 4 0 4 4, , 2, 2 .
2 2 2 2
10. False. The distance between the point (0, 0) and (4, 4) is
2 2 2 2( , ) (4 0) (4 0) 4 4
16 16 32 4 2
d P Q
11. Any three of the following: 2, 5 , 1,7 , 3, 9 , 5, 17 , 6, 21
12. Any three of the following: 3,3 , 5, 21 , 8,18 , 4,6 , 0, 6
13. Any three of the following: (1999, 35), (2001, 29), (2003, 22), (2005, 23), (2007, 20), (2009, 20)
14. Any three of the following:
2002,86.8 , 2004, 89.8 , 2006, 90.7 ,
2008, 97.4 , 2010, 106.5 , 2012,111.4 ,
2014, 111.5
15. P(–5, –6), Q(7, –1)
(a) 2 2
2 2
( , ) [7 – (–5)] [ 1 – (–6)]
12 5 169 13
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
–5 7 6 ( 1) 2 7, ,
2 2 2 27
1, .2
College Algebra and Trigonometry 6th Edition Lial Solutions ManualFull Download: http://testbanklive.com/download/college-algebra-and-trigonometry-6th-edition-lial-solutions-manual/
Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com
Section 2.1 Rectangular Coordinates and Graphs 171
Copyright © 2017 Pearson Education, Inc.
16. P(–4, 3), Q(2, –5)
(a) 2 2
2 2
( , ) [2 – (– 4)] (–5 – 3)
6 (–8) 100 10
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 4 2 3 (–5) 2 2, ,
2 2 2 21, 1 .
17. (8, 2), (3, 5)P Q
(a)
2 2
2 2
( , ) (3 – 8) (5 – 2)
5 3
25 9 34
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
8 3 2 5 11 7, ,
2 2 2 2
.
18. P (−8, 4), Q (3, −5)
(a) 2 2
2 2
( , ) 3 – 8 5 4
11 (–9) 121 81
202
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
–8 3 4 (–5) 5 1, , .
2 2 2 2
19. P(–6, –5), Q(6, 10)
(a) 2 2
2 2
( , ) [6 – (– 6)] [10 – (–5)]
12 15 144 225
369 3 41
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 6 6 –5 10 0 5 5, , 0, .
2 2 2 2 2
20. P(6, –2), Q(4, 6)
(a) 2 2
2 2
( , ) (4 – 6) [6 – (–2)]
(–2) 8
4 64 68 2 17
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
6 4 2 6 10 4, , 5, 2
2 2 2 2
21. 3 2, 4 5 ,P 2, – 5Q
(a)
2 2
2 2
( , )
2 – 3 2 – 5 – 4 5
–2 2 –5 5
8 125 133
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
3 2 2 4 5 (– 5),
2 2
4 2 3 5 3 5, 2 2, .
2 2 2
22. – 7, 8 3 , 5 7, – 3P Q
(a) 2 2
2 2
( , )
[5 7 – (– 7)] (– 3 – 8 3)
(6 7) (–9 3) 252 243
495 3 55
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 7 5 7 8 3 (– 3),
2 2
4 7 7 3 7 3, 2 7, .
2 2 2
23. Label the points A(–6, –4), B(0, –2), and C(–10, 8). Use the distance formula to find the length of each side of the triangle.
2 2
2 2
( , ) [0 – (– 6)] [–2 – (– 4)]
6 2 36 4 40
d A B
2 2
2 2
( , ) (–10 – 0) [8 – (–2)]
( 10) 10 100 100
200
d B C
2 2
2 2
( , ) [–10 – (– 6)] [8 – (– 4)]
(– 4) 12 16 144 160
d A C
Because 2 2 240 160 200 ,
triangle ABC is a right triangle.
172 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
24. Label the points A(–2, –8), B(0, –4), and C(–4, –7). Use the distance formula to find the length of each side of the triangle.
2 2
2 2
( , ) [0 – (–2)] [– 4 – (–8)]
2 4 4 16 20
d A B
2 2
2 2
( , ) (– 4 – 0) [–7 – (– 4)]
(– 4) (–3) 16 9
25 5
d B C
2 2
2 2
( , ) [– 4 – (–2)] [–7 – (–8)]
(–2) 1 4 1 5
d A C
Because 2 2 2( 5) ( 20) 5 20 25 5 , triangle ABC is a right triangle.
25. Label the points A(–4, 1), B(1, 4), and C(–6, –1).
2 2
2 2
( , ) [1 – (– 4)] (4 – 1)
5 3 25 9 34
d A B
2 2
2 2
( , ) (– 6 –1) (–1 – 4)
(–7) (–5) 49 25 74
d B C
2 2
2 2
( , ) [– 6 – (– 4)] (–1 –1)
(–2) (–2) 4 4 8
d A C
Because 2 2 2( 8) ( 34) ( 74) because
8 34 42 74, triangle ABC is not a right triangle.
26. Label the points A(–2, –5), B(1, 7), and C(3, 15).
2 2
2 2
( , ) [1 ( 2)] [7 ( 5)]
3 12 9 144 153
d A B
2 2
2 2
( , ) (3 1) (15 7)
2 8 4 64 68
d B C
2 2
2 2
( , ) [3 ( 2)] [15 ( 5)]
5 20 25 400 425
d A C
Because 22 2( 68) ( 153) 425 because
68 153 221 425 , triangle ABC is not a right triangle.
27. Label the points A(–4, 3), B(2, 5), and C(–1, –6).
2 2
2 2
( , ) 2 – –4 5 3
6 2 36 4 40
d A B
2 2
2 2
( , ) 1 2 6 5
(– 3) (–11)
9 121 130
d B C
2 2
22
( , ) –1 – –4 6 3
3 9 9 81 90
d A C
Because 2 2 240 90 130 , triangle
ABC is a right triangle.
28. Label the points A(–7, 4), B(6, –2), and C(0, –15).
2 2
22
( , ) 6 – –7 2 4
13 6
169 36 205
d A B
22
2 2
( , ) 0 6 –15 – –2
– 6 –13
36 169 205
d B C
2 2
22
( , ) 0 – –7 15 4
7 19 49 361 410
d A C
Because 2 2 2205 205 410 ,
triangle ABC is a right triangle.
29. Label the given points A(0, –7), B(–3, 5), and C(2, –15). Find the distance between each pair of points.
22
2 2
( , ) 3 0 5 – –7
–3 12 9 144
153 3 17
d A B
2 2
22
( , ) 2 – 3 –15 – 5
5 –20 25 400
425 5 17
d B C
22
22
( , ) 2 0 15 – 7
2 –8 68 2 17
d A C
Because ( , ) ( , ) ( , )d A B d A C d B C or
3 17 2 17 5 17, the points are collinear.
Section 2.1 Rectangular Coordinates and Graphs 173
Copyright © 2017 Pearson Education, Inc.
30. Label the points A(–1, 4), B(–2, –1), and C(1, 14). Apply the distance formula to each pair of points.
2 2
2 2
( , ) –2 – –1 –1 – 4
–1 –5 26
d A B
2 2
2 2
( , ) 1 – –2 14 – –1
3 15 234 3 26
d B C
2 2
2 2
( , ) 1 – –1 14 – 4
2 10 104 2 26
d A C
Because 26 2 26 3 26 , the points are collinear.
31. Label the points A(0, 9), B(–3, –7), and C(2, 19).
2 2
2 2
( , ) (–3 – 0) (–7 – 9)
(–3) (–16) 9 256
265 16.279
d A B
2 2
2 2
( , ) 2 – –3 19 – –7
5 26 25 676
701 26.476
d B C
2 2
2 2
( , ) 2 – 0 19 – 9
2 10 4 100
104 10.198
d A C
Because ( , ) ( , ) ( , )d A B d A C d B C
or 265 104 70116.279 10.198 26.476,
26.477 26.476,
the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)
32. Label the points A(–1, –3), B(–5, 12), and C(1, –11).
2 2
2 2
( , ) –5 – –1 12 – –3
– 4 15 16 225
241 15.5242
d A B
2 2
22
( , ) 1 – –5 –11 –12
6 –23 36 529
565 23.7697
d B C
2 2
22
( , ) 1 – –1 –11 – –3
2 –8 4 64
68 8.2462
d A C
Because d(A, B) + d(A, C) d(B, C)
or 241 68 56515.5242 8.2462 23.7697
23.7704 23.7697,
the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)
33. Label the points A(–7, 4), B(6,–2), and C(–1,1).
2 2
22
( , ) 6 – –7 2 4
13 6 169 36
205 14.3178
d A B
22
2 2
( , ) 1 6 1 –2
7 3 49 9
58 7.6158
d B C
2 2
22
( , ) 1 – –7 1 4
6 –3 36 9
45 6.7082
d A C
Because d(B, C) + d(A, C) d(A, B) or
58 45 2057.6158 6.7082 14.3178
14.3240 14.3178,
the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)
34. Label the given points A(–4, 3), B(2, 5), and C(–1, 4). Find the distance between each pair of points.
2 2 2 2( , ) 2 – –4 5 3 6 2
36 4 40 2 10
d A B
2 2
2 2
( , ) ( 1 2) (4 5)
3 (–1) 9 1 10
d B C
2 2
2 2
( , ) 1 – 4 4 3
3 1 9 1 10
d A C
Because ( , ) ( , ) ( , )d B C d A C d A B or
10 10 2 10, the points are collinear.
174 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
35. Midpoint (5, 8), endpoint (13, 10) 13 10
5 and 82 2
13 10 and 10 16
–3 and 6.
x y
x yx y
The other endpoint has coordinates (–3, 6).
36. Midpoint (–7, 6), endpoint (–9, 9) –9 9
–7 and 62 2
–9 –14 and 9 12
–5 and 3.
x y
x yx y
The other endpoint has coordinates (–5, 3).
37. Midpoint (12, 6), endpoint (19, 16) 19 16
12 and 62 2
19 24 and 16 12
5 and – 4.
x y
x yx y
The other endpoint has coordinates (5, –4).
38. Midpoint (–9, 8), endpoint (–16, 9) –16 9
–9 and 82 2
–16 –18 and 9 16
–2 and 7
x y
x yx y
The other endpoint has coordinates (–2, 7).
39. Midpoint (a, b), endpoint (p, q)
and2 2
2 and 2
2 and 2
p x q ya b
p x a q y bx a p y b q
The other endpoint has coordinates (2 , 2 )a p b q .
40. Midpoint 6 , 6a b , endpoint 3 , 5a b
3 56 and 6
2 23 12 and 5 12
9 and 7
a x b ya b
a x a b y bx a y b
The other endpoint has coordinates (9a, 7b).
41. The endpoints of the segment are (1990, 21.3) and (2012, 30.1).
1990 2012 21.3 30.9,
2 22001, 26.1
M
The estimate is 26.1%. This is very close to the actual figure of 26.2%.
42. The endpoints are (2006, 7505) and (2012, 3335)
2006 2012 7505 3335,
2 22009, 5420
M
According to the model, the average national advertising revenue in 2009 was $5420 million. This is higher than the actual value of $4424 million.
43. The points to use are (2011, 23021) and (2013, 23834). Their midpoint is
2011 2013 23,021 23,834,
2 2(2012, 23427.5).
In 2012, the poverty level cutoff was approximately $23,428.
44. (a) To estimate the enrollment for 2003, use the points (2000, 11,753) and (2006, 13,180)
2000 2006 11,753 13,180,
2 22003, 12466.5
M
The enrollment for 2003 was about 12,466.5 thousand.
(b) To estimate the enrollment for 2009, use the points (2006, 13,180) and (2012, 14,880)
2006 2012 13,180 14,880,
2 22009, 14030
M
The enrollment for 2009 was about 14,030 thousand.
45. The midpoint M has coordinates
1 2 1 2, .2 2
x x y y
2 21 2 1 2
1 1
2 21 2 1 1 2 1
2 22 1 2 1
2 22 1 2 1
2 22 1 2 1
2 212 1 2 12
( , )
– –2 2
2 2– –
2 2 2 2
2 2
4 4
4
d P M
x x y yx y
x x x y y y
x x y y
x x y y
x x y y
x x y y
(continued on next page)
Section 2.1 Rectangular Coordinates and Graphs 175
Copyright © 2017 Pearson Education, Inc.
(continued)
2 21 2 1 2
2 2
2 22 1 2 2 1 2
2 22 1 2 1
2 22 1 2 1
2 22 1 2 1
2 212 1 2 12
( , )
2 2
2 2
2 2 2 2
2 2
4 4
4
d M Q
x x y yx y
x x x y y y
x x y y
x x y y
x x y y
x x y y
2 22 1 2 1( , )d P Q x x y y
Because
2 212 1 2 12
2 212 1 2 12
2 22 1 2 1 ,
x x y y
x x y y
x x y y
this shows ( , ) ( , ) ( , )d P M d M Q d P Q and
( , ) ( , ).d P M d M Q
46. The distance formula, 2 2
2 1 2 1( – ) ( – ) ,d x x y y can be written
as 2 2 1/ 22 1 2 1[( – ) ( – ) ] .d x x y y
In exercises 47−58, other ordered pairs are possible.
47. (a) x y 0 −2 y-intercept:
12
00 2 2
xy
4 0 x-intercept:
1212
00 2
2 4
yxx x
2 −1 additional point
(b)
48. (a) x y 0 2 y-intercept:
0
10 2 2
2
x
y
4 0 x-intercept: 0
10 2
21
2 42
y
x
x x
2 1 additional point
(b)
49. (a) x y 0 5
3 y-intercept:
53
02 0 3 5
3 5
xy
y y
52
0 x-intercept:
52
02 3 0 5
2 5
yxx x
4 −1 additional point
(b)
176 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
50. (a) x y 0 −3 y-intercept:
0
3 0 2 62 6 3
xy
y y
2 0 x-intercept:
0
3 2 0 63 6 2
yxx x
4 3 additional point
(b)
51. (a) x y 0 0 x- and y-intercept:
20 0
1 1 additional point
−2 4 additional point
(b)
52. (a) x y 0 2 y-intercept:
2
0
0 20 2 2
xyy y
−1 3 additional point 2 6 additional point
no x-intercept: 2
20 0 2
2 2
y xx x
(b)
53. (a) x y 3 0 x-intercept:
0
0 30 3 3
yx
x x
4 1 additional point 7 2 additional point
no y-intercept:
0 0 3 3x y y
(b)
54. (a) x y 0 −3 y-intercept:
0
0 30 3 3
xyy y
4 −1 additional point
9 0 x-intercept: 0
0 3
3 9
yxx x
(b)
Section 2.1 Rectangular Coordinates and Graphs 177
Copyright © 2017 Pearson Education, Inc.
55. (a) x y 0 2 y-intercept:
00 2
2 2
xyy y
2 0 x-intercept: 0
0 20 2 2
yxx x
−2 4 additional point
4 2 additional point
(b)
56. (a) x y −2 −2 additional point −4 0 x-intercept:
00 4
0 40 4 4
yx
xx x
0 −4 y-intercept: 0
0 4
4 4
xyy y
(b)
57. (a) x y 0 0 x- and y-intercept:
30 0 −1 −1 additional point 2 8 additional point
(b)
58. (a) x y 0 0 x- and y-intercept:
30 0 1 −1 additional point2 −8 additional point
(b)
59. Points on the x-axis have y-coordinates equal to 0. The point on the x-axis will have the same x-coordinate as point (4, 3). Therefore, the line will intersect the x-axis at (4, 0).
60. Points on the y-axis have x-coordinates equal to 0. The point on the y-axis will have the same y-coordinate as point (4, 3). Therefore, the line will intersect the y-axis at (0, 3).
61. Because (a, b) is in the second quadrant, a is negative and b is positive. Therefore, (a, – b) will have a negative x–coordinate and a negative y-coordinate and will lie in quadrant III. (–a, b) will have a positive x-coordinate and a positive y-coordinateand will lie in quadrant I. (–a, – b) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. (b, a) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV.
178 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
62. Label the points ( 2, 2),A (13,10),B
(21, 5),C and (6, 13).D To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.
Use the distance formula to find the length of
each side.
2 2
2 2
2 2
22
( , ) 13 2 10 2
15 8 225 64
289 17
( , ) 21 13 5 10
8 15 64 225
289 17
d A B
d B C
22
2 2
( , ) 6 21 13 5
15 8
225 64 289 17
d C D
22
2 2
( , ) 2 6 2 13
8 15
64 225 289 17
d D A
Because all sides have equal length, the four points form a rhombus.
63. To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.
Use the distance formula to find the length of
each side.
2 2
2 2
( , ) 5 1 2 1
4 1 16 1 17
d A B
2 2
2 2
( , ) 3 5 4 2
2 2 4 4 8
d B C
2 2
2 2
( , ) 1 3 3 4
4 1
16 1 17
d C D
2 2
22
( , ) 1 1 1 3
2 2 4 4 8
d D A
Because d(A, B) = d(C, D) and d(B, C) = d(D, A), the points are the vertices of a parallelogram. Because d(A, B) ≠ d(B, C), the points are not the vertices of a rhombus.
64. For the points A(4, 5) and D(10, 14), the difference of the x-coordinates is 10 – 4 = 6 and the difference of the y-coordinates is 14 – 5 = 9. Dividing these differences by 3, we obtain 2 and 3, respectively. Adding 2 and 3 to the x and y coordinates of point A, respectively, we obtain B(4 + 2, 5 + 3) or B(6, 8).
Adding 2 and 3 to the x- and y- coordinates of point B, respectively, we obtain C(6 + 2, 8 + 3) or C(8, 11). The desired points are B(6, 8) and C(8, 11).
We check these by showing that d(A, B) = d(B, C) = d(C, D) and that d(A, D) = d(A, B) + d(B, C) + d(C, D).
2 2
2 2
2 2
2 2
( , ) 6 4 8 5
2 3 4 9 13
( , ) 8 6 11 8
2 3 4 9 13
d A B
d B C
2 2
2 2
2 2
2 2
( , ) 10 8 14 11
2 3 4 9 13
( , ) 10 4 14 5
6 9 36 81
117 9(13) 3 13
d C D
d A D
d(A, B), d(B, C), and d(C, D) all have the same measure and d(A, D) = d(A, B) + d(B, C) + d(C, D) Because
3 13 13 13 13.
Section 2.2 Circles 179
Copyright © 2017 Pearson Education, Inc.
Section 2.2 Circles
1. The circle with equation 2 2 49x y has center with coordinates (0, 0) and radius equal to 7.
2. The circle with center (3, 6) and radius 4 has
equation 23 6 16.x y
3. The graph of 2 24 7 9x y has center
with coordinates (4, –7).
4. The graph of 22 5 9x y has center with
coordinates (0, 5).
5. This circle has center (3, 2) and radius 5. This is graph B.
6. This circle has center (3, –2) and radius 5. This is graph C.
7. This circle has center (–3, 2) and radius 5. This is graph D.
8. This circle has center (–3, –2) and radius 5. This is graph A.
9. The graph of 2 2 0x y has center (0, 0) and radius 0. This is the point (0, 0). Therefore, there is one point on the graph.
10. 100 is not a real number, so there are no
points on the graph of 2 2 100.x y
11. (a) Center (0, 0), radius 6
2 2
2 2 2 2 2
0 0 6
0 0 6 36
x y
x y x y
(b)
12. (a) Center (0, 0), radius 9
2 2
2 2 2 2 2
0 0 9
0 0 9 81
x y
x y x y
(b)
13. (a) Center (2, 0), radius 6
2 2
2 2 2
2 2
2 0 6
2 0 6
( – 2) 36
x y
x yx y
(b)
14. (a) Center (3, 0), radius 3
2 2
2 2
3 0 3
3 9
x yx y
(b)
15. (a) Center (0, 4), radius 4
2 2
22
0 4 4
4 16
x yx y
(b)
180 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
16. (a) Center (0, –3), radius 7
22
22 2
2 2
0 3 7
0 3 7
( 3) 49
x y
x yx y
(b)
17. (a) Center (–2, 5), radius 4
2 2
2 2 2
2 2
2 5 4
[ – (–2)] ( – 5) 4
( 2) ( – 5) 16
x y
x yx y
(b)
18. (a) Center (4, 3), radius 5
2 2
2 2 2
2 2
4 3 5
4 3 5
4 3 25
x y
x yx y
(b)
19. (a) Center (5, –4), radius 7
22
2 2 2
2 2
5 4 7
( – 5) [ – (– 4)] 7
( – 5) ( 4) 49
x y
x yx y
(b)
20. (a) Center (–3, –2), radius 6
2 2
2 2 2
2 2
3 2 6
3 2 6
( 3) ( 2) 36
x y
x yx y
(b)
21. (a) Center 2, 2 , radius 2
2 2
2 2
2 2 2
2 2 2
x y
x y
(b)
22. (a) Center 3, 3 , radius 3
2 2
2 2 2
2 2
3 3 3
3 3 3
3 3 3
x y
x y
x y
Section 2.2 Circles 181
Copyright © 2017 Pearson Education, Inc.
(b)
23. (a) The center of the circle is located at the midpoint of the diameter determined by the points (1, 1) and (5, 1). Using the midpoint formula, we have
1 5 1 1, 3,1
2 2C
. The radius is
one-half the length of the diameter:
2 215 1 1 1 2
2r
The equation of the circle is
2 23 1 4x y
(b) Expand 2 23 1 4x y to find the
equation of the circle in general form:
2 2
2 2
2 2
3 1 46 9 2 1 4
6 2 6 0
x yx x y y
x y x y
24. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−1, 1) and (−1, −5).
Using the midpoint formula, we have
1 ( 1) 1 ( 5), 1, 2 .
2 2C
The radius is one-half the length of the diameter:
2 211 1 5 1 3
2r
The equation of the circle is
2 21 2 9x y
(b) Expand 2 21 2 9x y to find the
equation of the circle in general form:
2 2
2 2
2 2
1 2 92 1 4 4 9
2 4 4 0
x yx x y y
x y x y
25. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−2, 4) and (−2, 0). Using the midpoint formula, we have
2 ( 2) 4 0, 2, 2 .
2 2C
The radius is one-half the length of the diameter:
2 212 2 4 0 2
2r
The equation of the circle is
2 22 2 4x y
(b) Expand 2 22 2 4x y to find the
equation of the circle in general form:
2 2
2 2
2 2
2 2 44 4 4 4 4
4 4 4 0
x yx x y y
x y x y
26. (a) The center of the circle is located at the midpoint of the diameter determined by the points (0, −3) and (6, −3). Using the midpoint formula, we have
0 6 3 ( 3), 3, 3
2 2C
.
The radius is one-half the length of the diameter:
2216 0 3 3 3
2r
The equation of the circle is
2 23 3 9x y
(b) Expand 2 23 3 9x y to find the
equation of the circle in general form:
2 2
2 2
2 2
3 3 96 9 6 9 9
6 6 9 0
x yx x y y
x y x y
27. 2 2 6 8 9 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
6 8 –9
6 9 8 16 –9 9 16
3 4 16
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (–3, –4) and radius 4.
182 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
28. 2 2 8 – 6 16 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
8 – 6 16
8 16 – 6 9 –16 16 9
4 – 3 9
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (–4, 3) and radius 3.
29. 2 2 4 12 4x y x y Complete the square on x and y separately.
2 2
2 2
2 2
– 4 12 – 4
– 4 4 12 36 – 4 4 36
– 2 6 36
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (2, –6) and radius 6.
30. 2 2 –12 10 –25x y x y Complete the square on x and y separately.
2 2
2 2
2 2
–12 10 –25
–12 36 10 25
– 25 36 25
– 6 5 36
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (6, –5) and radius 6.
31. 2 24 4 4 –16 – 19 0x y x y Complete the square on x and y separately.
2 2
2 214
14
4 4 – 4 19
4 4 – 4 4
19 4 4 4
x x y y
x x y y
2 212
2 212
4 4 – 2 36
– 2 9
x y
x y
Yes, it is a circle with center 12
, 2 and
radius 3.
32. 2 29 9 12 – 18 – 23 0x y x y Complete the square on x and y separately.
2 243
2 24 43 9
49
9 9 – 2 23
9 9 – 2 1
23 9 9 1
x x y y
x x y y
2 223
2 223
9 9 –1 36
–1 4
x y
x y
Yes, it is a circle with center 23
, 1 and
radius 2.
33. 2 2 2 – 6 14 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
2 – 6 –14
2 1 – 6 9 –14 1 9
1 – 3 – 4
x x y y
x x y y
x y
The graph is nonexistent.
34. 2 2 4 – 8 32 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
4 – 8 –32
4 4 – 8 16
– 32 4 16
2 – 4 –12
x x y y
x x y y
x y
The graph is nonexistent.
35. 2 2 6 6 18 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
6 6 18
6 9 6 9 18 9 9
3 3 0
x x y y
x x y y
x y
The graph is the point (3, 3).
36. 2 2 4 4 8 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
4 4 8
4 4 4 4 8 4 4
2 2 0
x x y y
x x y y
x y
The graph is the point (−2, −2).
37. 2 29 9 6 6 23 0x y x y Complete the square on x and y separately.
2 2
2 22 23 3
22 2 232 1 2 1 1 13 9 3 9 9 9 9
2 2 225 51 13 3 9 3
9 6 9 6 23
9 9 23
x x y y
x x y y
x x y y
x y
Yes, it is a circle with center 1 13 3
, and
radius 53
.
Section 2.2 Circles 183
Copyright © 2017 Pearson Education, Inc.
38. 2 24 4 4 4 7 0x y x y Complete the square on x and y separately.
2 2
2 21 14 4
1 14 4
2 21 12 2
2 2 91 12 2 4
4 4 7
4 4 –
7 4 4
4 4 – 9
–
x x y y
x x y y
x y
x y
Yes, it is a circle with center 1 12 2
, and
radius 32
.
39. The equations of the three circles are 2 2( 7) ( 4) 25x y , 2 2( 9) ( 4) 169x y , and 2 2( 3) ( 9) 100x y . From the graph of the
three circles, it appears that the epicenter is located at (3, 1).
Check algebraically:
2 2
2 2
2 2
( 7) ( 4) 25
(3 7) (1 4) 25
4 3 25 25 25
x y
2 2
2 2
2 2
( 9) ( 4) 169
(3 9) (1 4) 169
12 5 169 169 169
x y
2 2
2 2
2 2
( 3) ( 9) 100
(3 3) (1 9) 100
6 ( 8) 100 100 100
x y
(3, 1) satisfies all three equations, so the epicenter is at (3, 1).
40. The three equations are 2 2( 3) ( 1) 5x y , 2 2( 5) ( 4) 36x y , and 2 2( 1) ( 4) 40x y . From the graph of the
three circles, it appears that the epicenter is located at (5, 2).
Check algebraically:
2 2
2 2
2 2
( 3) ( 1) 5
(5 3) (2 1) 5
2 1 5 5 5
x y
2 2
2 2
2
( 5) ( 4) 36
(5 5) (2 4) 36
6 36 36 36
x y
2 2
2 2
2 2
( 1) ( 4) 40
(5 1) (2 4) 40
6 ( 2) 40 40 40
x y
(5, 2) satisfies all three equations, so the epicenter is at (5, 2).
41. From the graph of the three circles, it appears that the epicenter is located at (−2, −2).
Check algebraically:
2 2
2 2
2 2
( 2) ( 1) 25
( 2 2) ( 2 1) 25
( 4) ( 3) 2525 25
x y
2 2
2 2
2 2
( 2) ( 2) 16
( 2 2) ( 2 2) 16
0 ( 4) 1616 16
x y
2 2
2 2
2 2
( 1) ( 2) 9
( 2 1) ( 2 2) 9
( 3) 0 99 9
x y
(−2, −2) satisfies all three equations, so the epicenter is at (−2, −2).
184 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
42. From the graph of the three circles, it appears that the epicenter is located at (5, 0).
Check algebraically:
2 2
2 2
2 2
( 2) ( 4) 25
(5 2) (0 4) 25
3 ( 4) 2525 25
x y
2 2
2 2
2 2
( 1) ( 3) 25
(5 1) (0 3) 25
4 3 2525 25
x y
2 2
2 2
2 2
( 3) ( 6) 100
(5 3) (0 6) 100
8 6 100100 100
x y
(5, 0) satisfies all three equations, so the epicenter is at (5, 0).
43. The radius of this circle is the distance from the center C(3, 2) to the x-axis. This distance is 2, so r = 2.
2 2 2
2 2
( – 3) ( – 2) 2
( – 3) ( – 2) 4
x yx y
44. The radius is the distance from the center C(–4, 3) to the point P(5, 8).
2 2
2 2
[5 – (– 4)] (8 – 3)
9 5 106
r
The equation of the circle is 2 2 2
2 3
[ – (– 4)] ( – 3) ( 106)
( 4) ( – 3) 106
x yx y
45. Label the points P(x, y) and Q(1, 3).
If ( , ) 4d P Q , 2 21 3 4x y
2 21 3 16.x y
If x = y, then we can either substitute x for y or y for x. Substituting x for y we solve the following:
2 2
2 2
2
2
2
1 3 16
1 2 9 6 16
2 8 10 16
2 8 6 0
4 3 0
x xx x x x
x xx xx x
To solve this equation, we can use the quadratic formula with a = 1, b = –4, and c = –3.
24 4 4 1 3
2 1
4 16 12 4 28
2 2
4 2 72 7
2
x
Because x = y, the points are
2 7, 2 7 and 2 – 7, 2 7 .
46. Let P(–2, 3) be a point which is 8 units from Q(x, y). We have
2 2( , ) 2 3 8d P Q x y
2 22 3 64.x y
Because x + y = 0, x = –y. We can either substitute x for y or y for x. Substituting
x for y we solve the following:
22
2 2
2 2
2
2
2 3 64
2 3 64
4 4 9 6 64
2 10 13 64
2 10 51 0
x x
x xx x x x
x xx x
To solve this equation, use the quadratic formula with a = 2, b = 10, and c = –51.
210 10 4 2 51
2 2
10 100 408
410 4 12710 508
4 410 2 127 5 127
4 2
x
Because y x the points are
5 127 5 127,
2 2
and
–5 127 5 127, .
2 2
Section 2.2 Circles 185
Copyright © 2017 Pearson Education, Inc.
47. Let P(x, y) be a point whose distance from
A(1, 0) is 10 and whose distance from
B(5, 4) is 10 . d(P, A) = 10 , so
2 2(1 ) (0 ) 10x y 2 2(1 ) 10.x y ( , ) 10, sod P B
2 2(5 ) (4 ) 10x y 2 2(5 ) (4 ) 10.x y Thus,
2 2 2 2
2 2
2 2
(1 ) (5 ) (4 )
1 2
25 10 16 81 2 41 10 8
8 40 85
x y x yx x y
x x y yx x yy xy x
Substitute 5 – x for y in the equation 2 2(1 ) 10x y and solve for x. 2 2
2 2
2 2
2
(1 ) (5 ) 10
1 2 25 10 10
2 12 26 10 2 12 16 0
6 8 0 ( 2)( 4) 02 0 or 4 0
2 or 4
x xx x x x
x x x xx x x xx x
x x
To find the corresponding values of y use the equation y = 5 – x. If x = 2, then y = 5 – 2 = 3. If x = 4, then y = 5 – 4 = 1. The points satisfying the conditions are (2, 3) and (4, 1).
48. The circle of smallest radius that contains the points A(1, 4) and B(–3, 2) within or on its boundary will be the circle having points A and B as endpoints of a diameter. The center will be M, the midpoint:
1 3 4 2 2 6, ,
2 2 2 2
(−1, 3).
The radius will be the distance from M to either A or B:
2 2
2 2
( , ) [1 ( 1)] (4 3)
2 1 4 1 5
d M A
The equation of the circle is
22 2
2 2
1 3 5
1 3 5.
x y
x y
49. Label the points A(3, y) and B(–2, 9). If d(A, B) = 12, then
2 2
2 2
2 2 2
2
2
2 3 9 12
5 9 12
5 9 12
25 81 18 144
18 38 0
y
y
yy y
y y
Solve this equation by using the quadratic formula with a = 1, b = –18, and c = –38:
218 18 4 1 38
2 1
18 324 152 18 476
2 1 2
y
18 4 119 18 2 119
9 1192 2
The values of y are 9 119 and 9 119 .
50. Because the center is in the third quadrant, the
radius is 2 , and the circle is tangent to both
axes, the center must be at ( 2, 2). Using the center-radius of the equation of a circle, we have
2 2 2
2 2
2 2 2
2 2 2.
x y
x y
51. Let P(x, y) be the point on the circle whose distance from the origin is the shortest. Complete the square on x and y separately to write the equation in center-radius form:
2 2
2 2
2 2
16 14 88 0
16 64 14 4988 64 49
( 8) ( 7) 25
x x y yx x y y
x y
So, the center is (8, 7) and the radius is 5.
2 2( , ) 8 7 113d C O . Because the
length of the radius is 5, ( , ) 113 5d P O .
186 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
52. Using compasses, draw circles centered at Wickenburg, Kingman, Phoenix, and Las Vegas with scaled radii of 50, 75, 105, and 180 miles respectively. The four circles should intersect at the location of Nothing.
53. The midpoint M has coordinates
3 –9–1+5 4 6 , , (2, – 3).
2 2 2 2
54. Use points C(2, –3) and P(–1, 3).
22
2 2
( , ) –1 – 2 3 – –3
–3 6 9 36
45 3 5
d C P
The radius is 3 5.
55. Use points C(2, –3) and Q(5, –9).
22
22
( , ) 5 – 2 –9 – –3
3 –6 9 36
45 3 5
d C Q
The radius is 3 5.
56. Use the points P(–1, 3) and Q(5, –9).
Because 2 2( , ) 5 – –1 –9 – 3d P Q
226 –12 36 144 180
6 5 ,the radius is 1
( , ).2
d P Q Thus
16 5 3 5.
2r
57. The center-radius form for this circle is 2 2 2( – 2) ( 3) (3 5)x y 2 2( – 2) ( 3) 45.x y
58. Label the endpoints of the diameter P(3, –5) and Q(–7, 3). The midpoint M of the segment joining P and Q has coordinates
3 (–7) –5 3 4 –2, , (–2, –1).
2 2 2 2
The center is C(–2, –1). To find the radius, we can use points C(–2, –1) and P(3, –5)
2 2
22
( , ) 3 – –2 –5 – –1
5 –4 25 16 41
d C P
We could also use points C(–2, –1).and Q(–7, 3).
2 2
2 2
( , ) 7 – –2 3 – –1
5 4 25 16 41
d C Q
We could also use points P(3, –5) and Q(–7, 3) to find the length of the diameter. The length of the radius is one-half the length of the diameter.
22
2 2
( , ) 7 – 3 3 – –5
10 8 100 64
164 2 41
d P Q
1 1( , ) 2 41 41
2 2d P Q
The center-radius form of the equation of the circle is
2 2 2
2 2
[ – (–2)] [ – (–1)] ( 41)
( 2) ( 1) 41
x yx y
59. Label the endpoints of the diameter P(–1, 2) and Q(11, 7). The midpoint M of the segment joining P and Q has coordinates
1 11 2 7 9, 5, .
2 2 2
The center is 92
5, .C To find the radius, we
can use points 92
5,C and P(−1, 2).
22 92
22 5 169 132 4 2
( , ) 5 – –1 2
6
d C P
We could also use points 92
5,C and
Q(11, 7).
22 92
22 5 169 132 4 2
( , ) 5 11 – 7
6
d C Q
(continued on next page)
Section 2.3 Functions 187
Copyright © 2017 Pearson Education, Inc.
(continued)
Using the points P and Q to find the length of the diameter, we have
2 2
2 2
, 1 11 2 7
12 5
169 13
d P Q
1 1 13, 13
2 2 2d P Q
The center-radius form of the equation of the circle is
2 22 9 132 2
22 9 1692 4
5
5
x y
x y
60. Label the endpoints of the diameter P(5, 4) and Q(−3, −2). The midpoint M of the segment joining P and Q has coordinates
5 ( 3) 4 ( 2), 1, 1 .
2 2
The center is C(1, 1). To find the radius, we can use points C(1, 1) and P(5, 4).
2 2
2 2
( , ) 5 1 4 1
4 3 25 5
d C P
We could also use points C(1, 1) and Q(−3, −2).
2 2
2 2
( , ) 1 3 1 –2
4 3 25 5
d C Q
Using the points P and Q to find the length of the diameter, we have
2 2
2 2
, 5 3 4 2
8 6 100 10
d P Q
1 1( , ) 10 5
2 2d P Q
The center-radius form of the equation of the circle is
2 2 2
2 2
1 1 5
1 1 25
x yx y
61. Label the endpoints of the diameter P(1, 4) and Q(5, 1). The midpoint M of the segment joining P and Q has coordinates
52
1 5 4 1, 3, .
2 2
The center is 52
3, .C
The length of the diameter PQ is
2 2 2 21 5 4 1 4 3 25 5.
The length of the radius is 512 2
5 .
The center-radius form of the equation of the circle is
2 22 5 52 2
22 5 252 4
3
3
x y
x y
62. Label the endpoints of the diameter P(−3, 10) and Q(5, −5). The midpoint M of the segment joining P and Q has coordinates
52
3 5 10 ( 5), 1, .
2 2
The center is 52
1, .C
The length of the diameter PQ is
22 2 23 5 10 5 8 15
289 17.
The length of the radius is 1712 2
17 .
The center-radius form of the equation of the circle is
2 22 5 172 2
22 5 2892 4
1
1
x y
x y
Section 2.3 Functions
1. The domain of the relation
3,5 , 4,9 , 10,13 is 3, 4,10 .
2. The range of the relation in Exercise 1 is
5,9,13 .
3. The equation y = 4x – 6 defines a function with independent variable x and dependent variable y.
4. The function in Exercise 3 includes the ordered pair (6, 18).
5. For the function 4 2,f x x
2 4 2 2 8 2 10.f
6. For the function ,g x x 9 9 3.g
7. The function in Exercise 6, ,g x x has
domain 0, .
188 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
8. The function in Exercise 6, ,g x x has
range 0, .
For exercises 9 and 10, use this graph.
9. The largest open interval over which the function graphed here increases is , 3 .
10. The largest open interval over which the function graphed here decreases is 3, .
11. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
12. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
13. Two ordered pairs, namely (2, 4) and (2, 6), have the same x-value paired with different y-values, so the relation is not a function.
14. Two ordered pairs, namely (9, −2) and (9, 1), have the same x-value paired with different y-values, so the relation is not a function.
15. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
16. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
17. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
18. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
19. Two sets of ordered pairs, namely (1, 1) and (1, −1) as well as (2, 4) and (2, −4), have the same x-value paired with different y-values, so the relation is not a function. domain: {0, 1, 2}; range: {−4, −1, 0, 1, 4}
20. The relation is not a function because the x-value 3 corresponds to two y-values, 7 and 9. This correspondence can be shown as follows.
domain: {2, 3, 5}; range: {5, 7, 9, 11}
21. The relation is a function because for each different x-value there is exactly one y-value. domain: {2, 3, 5, 11, 17}; range: {1, 7, 20}
22. The relation is a function because for each different x-value there is exactly one y-value. domain: {1, 2, 3, 5}; range: {10, 15, 19, 27}
23. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
Domain: {0, −1, −2}; range: {0, 1, 2}
Section 2.3 Functions 189
Copyright © 2017 Pearson Education, Inc.
24. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
Domain: {0, 1, 2}; range: {0, −1, −2}
25. The relation is a function because for each different year, there is exactly one number for visitors. domain: {2010, 2011, 2012, 2013} range: {64.9, 63.0, 65.1, 63.5}
26. The relation is a function because for each basketball season, there is only one number for attendance. domain: {2011, 2012, 2013, 2014} range: {11,159,999, 11,210,832, 11,339,285, 11,181,735}
27. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: ,
28. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: , 4
29. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: 3, ; range: ,
30. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: [−4, 4]; range: [−3, 3]
31. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: ,
32. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: [−2, 2]; range: [0, 4]
33. 2y x represents a function because y is always found by squaring x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the square of any real number is not negative, the range would be zero or greater.
domain: , ; range: 0,
34. 3y x represents a function because y is always found by cubing x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the cube of any real number could be negative, positive, or zero, the range would be any real number.
domain: , ; range: ,
35. The ordered pairs (1, 1) and (1, −1) both
satisfy 6.x y This equation does not represent a function. Because x is equal to the sixth power of y, the values of x are nonnegative. Any real number can be raised to the sixth power, so the range of the relation is all real numbers.
domain: 0, range: ,
190 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
36. The ordered pairs (1, 1) and (1, −1) both
satisfy 4.x y This equation does not represent a function. Because x is equal to the fourth power of y, the values of x are nonnegative. Any real number can be raised to the fourth power, so the range of the relation is all real numbers.
domain: 0, range: ,
37. 2 5y x represents a function because y is found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers.
domain: , ; range: ,
38. 6 4y x represents a function because y is found by multiplying x by −6 and adding 4. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is −6 times x, plus 4, y also may be any real number, and so the range is also all real numbers.
domain: , ; range: ,
39. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x + y < 3 corresponds to many values of y. The ordered pairs (1, –2), (1, 1), (1, 0), (1, −1), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.
domain: , ; range: ,
40. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x − y < 4 corresponds to many values of y. The ordered pairs (1, −1), (1, 0), (1, 1), (1, 2), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.
domain: , ; range: ,
Section 2.3 Functions 191
Copyright © 2017 Pearson Education, Inc.
41. For any choice of x in the domain of ,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x Because the radical is nonnegative, the range is also zero or greater.
domain: 0, ; range: 0,
42. For any choice of x in the domain of
,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x The outcome of the radical is nonnegative, when you change the sign (by multiplying by −1), the range becomes nonpositive. Thus the range is zero or less.
domain: 0, ; range: , 0
43. Because 2xy can be rewritten as 2 ,xy
we can see that y can be found by dividing x into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: , 0 0, ;
range: , 0 0,
44. Because xy = −6 can be rewritten as 6 ,xy
we can see that y can be found by dividing x into −6. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: , 0 0, ;
range: , 0 0,
45. For any choice of x in the domain of
4 1y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have
4 1 0 4 1x x 14
.x Because the
radical is nonnegative, the range is also zero or greater.
domain: 1
4, ; range: 0,
192 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
46. For any choice of x in the domain of
7 2y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have
7 2 0 2 7x x 7 72 2
or .x x
Because the radical is nonnegative, the range is also zero or greater.
domain: 72
, ; range: 0,
47. Given any value in the domain of 23
,xy we
find y by subtracting 3, then dividing into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 3. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: ,3 3, ;
range: , 0 0,
48. Given any value in the domain of 75
,xy we
find y by subtracting 5, then dividing into −7. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 5. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: ,5 5, ;
range: , 0 0,
49. B. The notation f(3) means the value of the dependent variable when the independent variable is 3.
50. Answers will vary. An example is: The cost of gasoline depends on the number of gallons used; so cost is a function of number of gallons.
51.
3 4
0 3 0 4 0 4 4
f x xf
52.
3 4
3 3 3 4 9 4 13
f x xf
53.
2
2
4 1
2 2 4 2 1
4 8 1 11
g x x xg
54.
2
2
4 1
10 10 4 10 1100 40 1 59
g x x xg
55. 1 1
3 3
3 4
3 4 1 4 3
f x xf
56. 7 7
3 3
3 4
3 4 7 4 11
f x xf
57.
2
21 1 12 2 2
1 114 4
4 1
4 1
2 1
g x x x
g
58.
2
21 1 14 4 4
1 116 16
4 1
4 1
1 1
g x x x
g
59.
3 4
3 4
f x xf p p
Section 2.3 Functions 193
Copyright © 2017 Pearson Education, Inc.
60.
2
2
4 1
4 1
g x x xg k k k
61.
3 4
3 4 3 4
f x xf x x x
62.
2
2
2
4 1
4 1
4 1
g x x xg x x x
x x
63.
3 4
2 3 2 43 6 4 3 2
f x xf x x
x x
64.
3 4
4 3 4 43 12 4 3 8
f x xf a a
a a
65.
3 4
2 3 3 2 3 46 9 4 6 13
f x xf m m
m m
66.
3 4
3 2 3 3 2 49 6 4 9 10
f x xf t t
t t
67. (a) 2 2f (b) 1 3f
68. (a) 2 5f (b) 1 11f
69. (a) 2 15f (b) 1 10f
70. (a) 2 1f (b) 1 7f
71. (a) 2 3f (b) 1 3f
72. (a) 2 3f (b) 1 2f
73. (a) 2 0f (b) 0 4f
(c) 1 2f (d) 4 4f
74. (a) 2 5f (b) 0 0f
(c) 1 2f (d) 4 4f
75. (a) 2 3f (b) 0 2f
(c) 1 0f (d) 4 2f
76. (a) 2 3f (b) 0 3f
(c) 1 3f (d) 4 3f
77. (a)
1 13 3
3 123 12
12
34 4
x yy x
xy
y x f x x
(b) 13
3 3 4 1 4 3f
78. (a)
1 14 4
4 88 4
8 48
42 2
x yx y
x yx y
y x f x x
(b) 3 3 8 514 4 4 4 4
3 3 2 2f
79. (a)
2
2
2
2 3
2 3
2 3
y x xy x x
f x x x
(b) 23 2 3 3 32 9 3 3 18
f
80. (a)
2
2
2
3 2
3 2
3 2
y x xy x x
f x x x
(b) 23 3 3 3 23 9 3 2 32
f
81. (a)
8 84 43 3 3 3
4 3 84 3 8
4 8 34 8
3
x yx y
x yx y
y x f x x
(b) 8 84 12 43 3 3 3 3
3 3f
82. (a)
9 92 25 5 5 5
2 5 95 2 9
2 9
5
x yy x
xy
y x f x x
(b) 9 6 9 1525 5 5 5 5
3 3 3f
83. 3 4f
194 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
84. Because 20.2 0.2 3 0.2 1f
= 0.04 + 0.6 + 1 = 1.64, the height of the rectangle is 1.64 units. The base measures 0.3 − 0.2 = 0.1 unit. Because the area of a rectangle is base times height, the area of this rectangle is 0.1(1.64) = 0.164 square unit.
85. 3f is the y-component of the coordinate,
which is −4.
86. 2f is the y-component of the coordinate,
which is −3.
87. (a) 2, 0 (b) , 2
(c) 0,
88. (a) 3, 1 (b) 1,
(c) , 3
89. (a) , 2 ; 2,
(b) 2, 2 (c) none
90. (a) 3,3 (b) , 3 ; 3,
(c) none
91. (a) 1, 0 ; 1,
(b) , 1 ; 0, 1
(c) none
92. (a) , 2 ; 0, 2
(b) 2, 0 ; 2,
(c) none
93. (a) Yes, it is the graph of a function.
(b) [0, 24]
(c) When t = 8, y = 1200 from the graph. At 8 A.M., approximately 1200 megawatts is being used.
(d) The most electricity was used at 17 hr or 5 P.M. The least electricity was used at 4 A.M.
(e) 12 1900f
At 12 noon, electricity use is about 1900 megawatts.
(f) increasing from 4 A.M. to 5 P.M.; decreasing from midnight to 4 A.M. and from 5 P.M. to midnight
94. (a) At t = 2, y = 240 from the graph. Therefore, at 2 seconds, the ball is 240 feet high.
(b) At y = 192, x = 1 and x = 5 from the graph. Therefore, the height will be 192 feet at 1 second and at 5 seconds.
(c) The ball is going up from 0 to 3 seconds and down from 3 to 7 seconds.
(d) The coordinate of the highest point is (3, 256). Therefore, it reaches a maximum height of 256 feet at 3 seconds.
(e) At x = 7, y = 0. Therefore, at 7 seconds, the ball hits the ground.
95. (a) At t = 12 and t = 20, y = 55 from the graph. Therefore, after about 12 noon until about 8 P.M. the temperature was over 55º.
(b) At t = 6 and t = 22, y = 40 from the graph. Therefore, until about 6 A.M. and after 10 P.M. the temperature was below 40º.
(c) The temperature at noon in Bratenahl, Ohio was 55º. Because the temperature in Greenville is 7º higher, we are looking for the time at which Bratenahl, Ohio was at or above 48º. This occurred at approximately 10 A.M and 8:30 P.M.
(d) The temperature is just below 40° from midnight to 6 A.M., when it begins to rise until it reaches a maximum of just below 65° at 4 P.M. It then begins to fall util it reaches just under 40° again at midnight.
96. (a) At t = 8, y = 24 from the graph. Therefore, there are 24 units of the drug in the bloodstream at 8 hours.
(b) The level increases between 0 and 2 hours after the drug is taken and decreases between 2 and 12 hours after the drug is taken.
(c) The coordinates of the highest point are (2, 64). Therefore, at 2 hours, the level of the drug in the bloodstream reaches its greatest value of 64 units.
(d) After the peak, y = 16 at t = 10. 10 hours – 2 hours = 8 hours after the peak. 8 additional hours are required for the level to drop to 16 units.
Section 2.4 Linear Functions 195
Copyright © 2017 Pearson Education, Inc.
(e) When the drug is administered, the level is 0 units. The level begins to rise quickly for 2 hours until it reaches a maximum of 64 units. The level then begins to decrease gradually until it reaches a level of 12 units, 12 hours after it was administered.
Section 2.4 Linear Functions
1. B; ( )f x = 3x + 6 is a linear function with y-intercept (0, 6).
2. H; x = 9 is a vertical line.
3. C; ( )f x = −8 is a constant function.
4. G; 2x – y = –4 or y = 2x + 4 is a linear equation with x-intercept (–2, 0) and y-intercept (0, 4).
5. A; ( )f x = 5x is a linear function whose graph passes through the origin, (0, 0). f(0) = 5(0) = 0.
6. D; 2( )f x x is a function that is not linear.
7. –3m matches graph C because the line falls rapidly as x increases.
8. m = 0 matches graph A because horizontal lines have slopes of 0.
9. m = 3 matches graph D because the line rises rapidly as x increases.
10. m is undefined for graph B because vertical lines have undefined slopes.
11. ( ) – 4f x x Use the intercepts.
(0) 0 – 4 – 4 :f y-intercept
0 – 4 4 :x x x-intercept Graph the line through (0, –4) and (4, 0).
The domain and range are both , .
12. ( ) – 4f x x Use the intercepts.
(0) –0 4 4 :f y-intercept
0 – 4 4 :x x x-intercept Graph the line through (0, 4) and (4, 0).
The domain and range are both , .
13. 12
( ) – 6f x x
Use the intercepts.
12
(0) 0 – 6 – 6 :f y-intercept
1 12 2
0 – 6 6 12 :x x x x-intercept
Graph the line through (0, –6) and (12, 0).
The domain and range are both , .
14. 23
( ) 2f x x
Use the intercepts.
23
(0) 0 2 2 :f y-intercept
2 23 3
0 2 –2 3 : -interceptx x x x
Graph the line through (0, 2) and (–3, 0).
The domain and range are both , .
196 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
15. ( ) 3f x x The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 1, y = 3(1) = 3 . Another point is (1, 3). Graph the line through (0, 0) and (1, 3).
The domain and range are both , .
16. –2f x x
The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 3, y = –2(3) = –6, so another point is (3, –6). Graph the line through (0, 0) and (3, –6).
The domain and range are both , .
17. ( ) – 4f x is a constant function.
The graph of ( ) 4f x is a horizontal line with a y-intercept of –4.
domain: , ; range: {–4}
18. ( ) 3f x is a constant function whose graph is a horizontal line with y-intercept of 3.
domain: , ; range: {3}
19. ( ) 0f x is a constant function whose graph is the x-axis.
domain: , ; range: {0}
20. 9f x x
The domain and range are both , .
21. 4 3 12x y Use the intercepts.
4 0 3 12 3 124 : -intercept
y yy y
4 3 0 12 4 123 : -intercept
x xx x
Graph the line through (0, 4) and (−3, 0).
The domain and range are both , .
22. 2 5 10;x y Use the intercepts.
2 0 5 10 5 102 : -intercept
y yy y
2 5 0 10 2 105 : -intercept
x xx x
Graph the line through (0, 2) and (5, 0):
The domain and range are both , .
Section 2.4 Linear Functions 197
Copyright © 2017 Pearson Education, Inc.
23. 3 4 0y x Use the intercepts.
3 4 0 0 3 0 0 : -intercepty y y y
3 0 4 0 4 0 0 : -interceptx x x x
The graph has just one intercept. Choose an additional value, say 3, for x.
3 4 3 0 3 12 03 12 4
y yy y
Graph the line through (0, 0) and (3, 4):
The domain and range are both , .
24. 3 2 0x y Use the intercepts. 3 0 2 0 2 0 0 : -intercepty y y y
3 2 0 0 3 0 0 :x x x x-intercept
The graph has just one intercept. Choose an additional value, say 2, for x.
3 2 2 0 6 2 02 6 3
y yy y
Graph the line through (0, 0) and (2, −3):
The domain and range are both , .
25. x = 3 is a vertical line, intersecting the x-axis at (3, 0).
domain: {3}; range: ,
26. x = –4 is a vertical line intersecting the x-axis at (–4, 0).
domain: {−4}; range: ,
27. 2x + 4 = 0 2 4 2x x is a vertical line intersecting the x-axis at (–2, 0).
domain: {−2}; range: ,
28. 3 6 0 –3 6 2x x x is a vertical
line intersecting the x-axis at (2, 0).
domain: {2}; range: ,
29. 5 0 5x x is a vertical line intersecting the x-axis at (5, 0).
domain: {5}; range: ,
198 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
30. 3 0 3x x is a vertical line
intersecting the x-axis at 3,0 .
domain: {−3}; range: ,
31. y = 5 is a horizontal line with y-intercept 5. Choice A resembles this.
32. y = –5 is a horizontal line with y-intercept –5. Choice C resembles this.
33. x = 5 is a vertical line with x-intercept 5. Choice D resembles this.
34. x = –5 is a vertical line with x-intercept –5. Choice B resembles this.
35.
36.
37.
38.
39. The rise is 2.5 feet while the run is 10 feet so
the slope is 2.5 110 4
0.25 25% . So A =
0.25, 2.510
C , D = 25%, and 14
E are all
expressions of the slope.
40. The pitch or slope is 14
. If the rise is 4 feet
then 1 rise 4
4 run x or x = 16 feet. So 16 feet in
the horizontal direction corresponds to a rise of 4 feet.
41. Through (2, –1) and (–3, –3) Let 1 1 22, –1, –3, and x y x 2 –3.y
Then rise –3 – (–1) –2y and
run –3 – 2 –5.x
The slope is rise –2 2
.run –5 5
ymx
42. Through (−3, 4) and (2, −8) Let 1 1 2 23, 4, 2, and –8.x y x y
Then rise –8 – 4 –12y and
run 2 – ( 3) 5.x
The slope is rise –12 12
.run 5 5
ymx
43. Through (5, 8) and (3, 12) Let 1 1 2 25, 8, 3, and 12.x y x y
Then rise 12 8 4y and
run 3 5 2.x
The slope is rise 4
2.run 2
ymx
44. Through (5, –3) and (1, –7)
1 1 2 2Let 5, –3, 1, and –7.x y x y
Then rise –7 – (–3) – 4y and
run 1 – 5 – 4.x
The slope is 4
1.4
ymx
45. Through (5, 9) and (–2, 9)
2 1
2 1
9 – 9 00
–2 5 –7
y yymx x x
46. Through (–2, 4) and (6, 4)
2 1
2 1
4 4 00
6 (–2) 8
y yymx x x
47. Horizontal, through (5, 1) The slope of every horizontal line is zero, so m = 0.
48. Horizontal, through (3 , 5) The slope of every horizontal line is zero, so m = 0.
49. Vertical, through (4, –7) The slope of every vertical line is undefined; m is undefined.
Section 2.4 Linear Functions 199
Copyright © 2017 Pearson Education, Inc.
50. Vertical, through (–8, 5) The slope of every vertical line is undefined; m is undefined.
51. (a) 3 5y x Find two ordered pairs that are solutions to the equation. If x = 0, then 3 0 5 5.y y
If x = −1, then 3 1 5 3 5 2.y y y
Thus two ordered pairs are (0, 5) and (−1, 2)
2 1
2 1
rise 2 5 33.
run 1 0 1
y ymx x
(b)
52. 2 4y x Find two ordered pairs that are solutions to the equation. If 0,x then 2 0 4y
4.y If 1,x then 2 1 4y
2 4 2.y y Thus two ordered pairs
are 0, 4 and 1, 2 .
2 1
2 1
2 4rise 22.
run 1 0 1
y ymx x
(b)
53. 2 3y x Find two ordered pairs that are solutions to the equation. If 0,x then 2 0 0.y y
If 3,y then 2 3 3 6 3x x
2.x Thus two ordered pairs are 0,0 and
2, 3 .
2 1
2 1
rise 3 0 3.
run 2 0 2
y ymx x
(b)
54. 4 5y x Find two ordered pairs that are solutions to the equation. If 0,x then 4 0 0.y y
If 4,x then 4 5 4 4 20y y
5.y Thus two ordered pairs are 0,0 and
4, 5 .
2 1
2 1
rise 5 0 5.
run 4 0 4
y ymx x
(b)
55. 5 2 10x y Find two ordered pairs that are solutions to the equation. If 0,x then 5 0 2 10y
5.y If 0,y then 5 2 0 10x
5 10x 2.x
Thus two ordered pairs are 0, 5 and 2,0 .
2 1
2 1
0 5rise 5.
run 2 0 2
y ymx x
(b)
200 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
56. 4 3 12x y Find two ordered pairs that are solutions to the equation. If 0,x then 4 0 3 12y
3 12y 4.y If 0,y then
4 3 0 12 4 12x x 3.x Thus two
ordered pairs are 0, 4 and 3,0 .
2 1
2 1
rise 0 4 4.
run 3 0 3
y ymx x
(b)
57. Through (–1, 3), 32
m
First locate the point (–1, 3). Because the
slope is 32
, a change of 2 units horizontally (2
units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (1, 6), which can be used to complete the graph.
58. Through (–2, 8), 25
m . Because the slope is
25
, a change of 5 units horizontally (to the
right) produces a change of 2 units vertically (2 units up). This gives a second point (3, 10), which can be used to complete the graph. Alternatively, a change of 5 units to the left produces a change of 2 units down. This gives the point (−7, 6).
59. Through (3, –4), 13
–m . First locate the point
(3, –4). Because the slope is 13
– , a change of
3 units horizontally (3 units to the right) produces a change of –1 unit vertically (1 unit down). This gives a second point, (6, –5), which can be used to complete the graph.
60. Through (–2, –3), 34
–m . Because the slope
is 3 –34 4
– , a change of 4 units horizontally
(4 units to the right) produces a change of –3 units vertically (3 units down). This gives a second point (2, –6), which can be used to complete the graph.
61. Through 12
, 4 , m = 0.
The graph is the horizontal line through
12
, 4 .
Exercise 61 Exercise 62
62. Through 32
, 2 , m = 0.
The graph is the horizontal line through
32
, 2 .
Section 2.4 Linear Functions 201
Copyright © 2017 Pearson Education, Inc.
63. Through 52
, 3 , undefined slope. The slope
is undefined, so the line is vertical,
intersecting the x-axis at 52
, 0 .
64. Through 94
, 2 , undefined slope. The slope is
undefined, so the line is vertical, intersecting
the x-axis at 94
, 0 .
65. The average rate of change is f b f a
mb a
20 4 16$4
0 4 4
(thousand) per year. The
value of the machine is decreasing $4000 each year during these years.
66. The average rate of change is f b f a
mb a
200 0 200$50
4 0 4
per month. The amount
saved is increasing $50 each month during these months.
67. The graph is a horizontal line, so the average rate of change (slope) is 0. The percent of pay raise is not changing—it is 3% each year.
68. The graph is a horizontal line, so the average rate of change (slope) is 0. That means that the number of named hurricanes remained the same, 10, for the four consecutive years shown.
69. 2562 5085 2523
2012 1980 3278.8 thousand per year
f b f am
b a
The number of high school dropouts decreased by an average of 78.8 thousand per year from 1980 to 2012.
70. 1709 5302
2013 20063593
$513.297
f b f am
b a
Sales of plasma flat-panel TVs decreased by an average of $513.29 million per year from 2006 to 2013.
71. (a) The slope of –0.0167 indicates that the average rate of change of the winning time for the 5000 m run is 0.0167 min less. It is negative because the times are generally decreasing as time progresses.
(b) The Olympics were not held during World Wars I (1914−1919) and II (1939–1945).
(c) 0.0167 2000 46.45 13.05 miny
The model predicts a winning time of 13.05 minutes. The times differ by 13.35 − 13.05 = 0.30 min.
72. (a) From the equation, the slope is 200.02. This means that the number of radio stations increased by an average of 200.02 per year.
(b) The year 2018 is represented by x = 68. 200.02 68 2727.7 16,329.06y
According to the model, there will be about 16,329 radio stations in 2018.
73. 2013 2008 335,652 270,334
2013 2008 2013 200865,318
13,063.65
f f
The average annual rate of change from 2008 through 2013 is about 13,064 thousand.
74. 2014 2006 3.74 4.53
2014 2006 2014 20060.79
0.0998
f f
The average annual rate of change from 2006 through 2014 is about –0.099.
202 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
75. (a) 56.3 138
2013 200381.7
8.1710
f b f am
b a
The average rate of change was −8.17 thousand mobile homes per year.
(b) The negative slope means that the number of mobile homes decreased by an average of 8.17 thousand each year from 2003 to 2013.
76. 2013 1991 26.6 61.8
2013 1991 2013 199135.2
1.622
f f
There was an average decrease of 1.6 births per thousand per year from 1991 through 2013.
77. (a) 10 500C x x
(b) 35R x x
(c) 35 10 500
35 10 500 25 500
P x R x C xx xx x x
(d) 10 500 35
500 2520
C x R xx x
xx
20 units; do not produce
78. (a) 150 2700C x x
(b) 280R x x
(c) 280 150 2700
280 150 2700130 2700
P x R x C xx xx xx
(d) 150 2700 280
2700 13020.77 or 21 units
C x R xx x
xx
21 units; produce
79. (a) 400 1650C x x
(b) 305R x x
(c) 305 400 1650
305 400 165095 1650
P x R x C xx xx xx
(d) 400 1650 305
95 1650 095 1650
17.37 units
C x R xx xx
xx
This result indicates a negative “break-even point,” but the number of units produced must be a positive number. A calculator graph of the lines
1 400 1650y C x x and
2 305y R x x in the window
[0, 70] × [0, 20000] or solving the inequality 305 400 1650x x will show
that R x C x for all positive values
of x (in fact whenever x is greater than −17.4). Do not produce the product because it is impossible to make a profit.
80. (a) 11 180C x x
(b) 20R x x
(c) 20 11 180
20 11 180 9 180
P x R x C xx xx x x
(d) 11 180 20
180 920
C x R xx x
xx
20 units; produce
81. 200 1000 2401000 40 25C x R x x x
x x
The break-even point is 25 units. (25) 200 25 1000 $6000C which is the
same as (25) 240 25 $6000R
C x
R x
Chapter 2 Quiz (Sections 2.1–2.4) 203
Copyright © 2017 Pearson Education, Inc.
82. 220 1000 2401000 20 50C x R x x x
x x
The break-even point is 50 units instead of 25 units. The manager is not better off because twice as many units must be sold before beginning to show a profit.
83. The first two points are A(0, –6) and B(1, –3). 3 – (–6) 3
31 – 0 1
m
84. The second and third points are B(1, –3) and C(2, 0).
0 – (–3) 33
2 – 1 1m
85. If we use any two points on a line to find its slope, we find that the slope is the same in all cases.
86. The first two points are A(0, –6) and B(1, –3). 2 2
2 2
( , ) (1 – 0) [–3 – (– 6)]
1 3 1 9 10
d A B
87. The second and fourth points are B(1, –3) and D(3, 3).
2 2
2 2
( , ) (3 – 1) [3 – (–3)]
2 6 4 36
40 2 10
d B D
88. The first and fourth points are A(0, –6) and D(3, 3).
2 2
2 2
( , ) (3 – 0) [3 – (– 6)]
3 9 9 81
90 3 10
d A D
89. 10 2 10 3 10; The sum is 3 10, which is equal to the answer in Exercise 88.
90. If points A, B, and C lie on a line in that order, then the distance between A and B added to the distance between B and C is equal to the distance between A and C.
91. The midpoint of the segment joining A(0, –6) and G(6, 12) has coordinates
0 6 6 122 2
, 6 62 2
, 3,3 . The midpoint is
M(3, 3), which is the same as the middle entry in the table.
92. The midpoint of the segment joining E(4, 6) and F(5, 9) has coordinates
4 5 6 9 9 152 2 2 2
, , = (4.5, 7.5). If the
x-value 4.5 were in the table, the corresponding y-value would be 7.5.
Chapter 2 Quiz (Sections 2.1−2.4)
1.
2 22 1 2 1
2 2
2 2
( , )
8 ( 4) 3 2
( 4) ( 5) 16 25 41
d A B x x y y
2. To find an estimate for 2006, find the midpoint of (2004, 6.55) and (2008, 6.97:
2004 2008 6.55 6.97,
2 22006, 6.76
M
The estimated enrollment for 2006 was 6.76 million. To find an estimate for 2010, find the midpoint of (2008, 6.97) and (2012, 7.50):
2008 2012 6.97 7.50,
2 22010, 7.235
M
The estimated enrollment for 2006 was about 7.24 million.
3.
4.
5. 2 2 4 8 3 0x y x y Complete the square on x and y separately.
2 2
2 2
( 4 4) ( 8 16) 3 4 16
( 2) ( 4) 17
x x y yx y
The radius is 17 and the midpoint of the circle is (2, −4).
6. From the graph, f(−1) is 2.
7. Domain: ( , ); range: [0, )
204 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
8. (a) The largest open interval over which f is decreasing is ( , 3) .
(b) The largest open interval over which f is increasing is 3, .
(c) There is no interval over which the function is constant.
9. (a) 11 5 6 3
5 1 4 2m
(b) 4 4 0
01 ( 7) 6
m
(c) 4 12 16
6 6 0m
the slope is
undefined.
10. The points to use are (2009, 10,602) and (2013, 15,884). The average rate of change is
15,884 10,602 5282
1320.52013 2009 4
The average rate of change was 1320.5 thousand cars per year. This means that the number of new motor vehicles sold in the United States increased by an average of 1320.5 thousand per year from 2009 to 2013.
Section 2.5 Equations of Lines and Linear Models
1. The graph of the line 3 4 8y x has
slope 4 and passes through the point (8, 3).
2. The graph of the line 2 7y x has slope –2 and y-intercept (0, 7).
3. The vertical line through the point (–4, 8) has equation x = –4.
4. The horizontal line through the point (–4, 8) has equation y = 8.
For exercises 5 and 6, 67
6 7 0 7 6x y y x y x
5. Any line parallel to the graph of 6 7 0x y
must have slope 67.
6. Any line perpendicular to the graph of
6 7 0x y must have slope 76.
7. 14
2y x is graphed in D.
The slope is 14
and the y-intercept is (0, 2).
8. 4x + 3y = 12 or 3y = –4x + 12 43
or – 4y x
is graphed in B. The slope is 43
– and the
y-intercept is (0, 4).
9. 32
– 1 1y x is graphed in C. The slope
is 32
and a point on the graph is (1, –1).
10. y = 4 is graphed in A. y = 4 is a horizontal line with y-intercept (0, 4).
11. Through (1, 3), m = –2. Write the equation in point-slope form.
1 1– – 3 –2 –1y y m x x y x
Then, change to standard form. 3 –2 2 2 5y x x y
12. Through (2, 4), m = –1 Write the equation in point-slope form.
1 1– – 4 –1 – 2y y m x x y x
Then, change to standard form. – 4 – 2 6y x x y
13. Through (–5, 4), 32
m
Write the equation in point-slope form.
32
– 4 5y x
Change to standard form. 2 – 4 –3 52 – 8 –3 – 15
3 2 –7
y xy x
x y
14. Through 34
(– 4, 3), m
Write the equation in point-slope form.
34
– 3 4y x
Change to standard form. 4 3 3 4
4 12 3 123 4 24 or 3 4 24
y xy xx y x y
15. Through (–8, 4), undefined slope Because undefined slope indicates a vertical line, the equation will have the form x = a. The equation of the line is x = –8.
16. Through (5, 1), undefined slope This is a vertical line through (5, 1), so the equation is x = 5.
17. Through (5, −8), m = 0 This is a horizontal line through (5, −8), so the equation is y = −8.
18. Through (−3, 12), m = 0 This is a horizontal line through (−3, 12), so the equation is y = 12.
Section 2.5 Equations of Lines and Linear Models 205
Copyright © 2017 Pearson Education, Inc.
19. Through (–1, 3) and (3, 4) First find m.
4 – 3 1
3 – (–1) 4m
Use either point and the point-slope form.
14
– 4 – 34 16 3
4 134 13
y xy xx yx y
20. Through (2, 3) and (−1, 2) First find m.
2 3 1 1
1 2 3 3m
Use either point and the point-slope form.
13
3 – 23 – 9 2
3 73 7
y xy x
x yx y
21. x-intercept (3, 0), y-intercept (0, –2) The line passes through (3, 0) and (0, –2). Use these points to find m.
–2 – 0 2
0 – 3 3m
Using slope-intercept form we have 23
– 2.y x
22. x-intercept (–4, 0), y-intercept (0, 3) The line passes through the points (–4, 0) and (0, 3). Use these points to find m.
3 – 0 3
0 – (–4) 4m
Using slope-intercept form we have 34
3.y x
23. Vertical, through (–6, 4) The equation of a vertical line has an equation of the form x = a. Because the line passes through (–6, 4), the equation is x = –6. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)
24. Vertical, through (2, 7) The equation of a vertical line has an equation of the form x = a. Because the line passes through (2, 7), the equation is x = 2. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)
25. Horizontal, through (−7, 4) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−7, 4), the equation is y = 4.
26. Horizontal, through (−8, −2) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−8, −2), the equation is y = −2.
27. m = 5, b = 15 Using slope-intercept form, we have
5 15.y x
28. m = −2, b = 12 Using slope-intercept form, we have
2 12.y x
29. Through (−2, 5), slope = −4
5 4 2
5 4 25 4 8
4 3
y xy xy x
y x
30. Through (4, −7), slope = −2 7 2 4
7 2 82 1
y xy x
y x
31. slope 0, y-intercept 32
0,
These represent 32
0 and .m b
Using slope-intercept form, we have
3 32 2
0 .y x y
32. slope 0, y-intercept 54
0,
These represent 54
0 and .m b
Using slope-intercept form, we have
5 54 4
0 .y x y
33. The line x + 2 = 0 has x-intercept (–2, 0). It does not have a y-intercept. The slope of his line is undefined. The line 4y = 2 has
y-intercept 12
0, . It does not have an
x-intercept. The slope of this line is 0.
34. (a) The graph of y = 3x + 2 has a positive slope and a positive y-intercept. These conditions match graph D.
206 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b) The graph of y = –3x + 2 has a negative slope and a positive y-intercept. These conditions match graph B.
(c) The graph of y = 3x – 2 has a positive slope and a negative y-intercept. These conditions match graph A.
(d) The graph of y = –3x – 2 has a negative slope and a negative y-intercept. These conditions match graph C.
35. y = 3x – 1 This equation is in the slope-intercept form, y = mx + b. slope: 3; y-intercept: (0, –1)
36. y = –2x + 7 slope: –2; y-intercept: (0, 7)
37. 4x – y = 7 Solve for y to write the equation in slope-intercept form. – – 4 7 4 – 7y x y x slope: 4; y-intercept: (0, –7)
38. 2x + 3y = 16 Solve the equation for y to write the equation in slope-intercept form.
1623 3
3 –2 16 –y x y x
slope: 23
– ; y-intercept: 163
0,
39. 3 34 4
4 –3 – or 0y x y x y x
slope: 34
; y-intercept (0, 0)
40. 1 12 2
2 or 0y x y x y x
slope is 12
; y-intercept: (0, 0)
41. 2 4x y Solve the equation for y to write the equation in slope-intercept form.
12
2 – 4 – 2y x y x
slope: 12
– ; y-intercept: (0, −2)
Section 2.5 Equations of Lines and Linear Models 207
Copyright © 2017 Pearson Education, Inc.
42. 3 9x y Solve the equation for y to write the equation in slope-intercept form.
13
3 – 9 – 3y x y x
slope: 13
– ; y-intercept: (0,−3)
43. 32
1 0y x
Solve the equation for y to write the equation in slope-intercept form.
3 32 2
1 0 1y x y x
slope: 32
; y-intercept: (0, 1)
44. (a) Use the first two points in the table, A(–2, –11) and B(–1, –8).
–8 – (–11) 33
–1 – (–2) 1m
(b) When x = 0, y = −5. The y-intercept is (0, –5).
(c) Substitute 3 for m and –5 for b in the slope-intercept form.
3 – 5y mx b y x
45. (a) The line falls 2 units each time the x value increases by 1 unit. Therefore the slope is −2. The graph intersects the y-axis at the point (0, 1) and intersects the
x-axis at 12
, 0 , so the y-intercept is
(0, 1) and the x-intercept is 12
, 0 .
(b) An equation defining f is y = −2x + 1.
46. (a) The line rises 2 units each time the x value increases by 1 unit. Therefore the slope is 2. The graph intersects the y-axis at the point (0, −1) and intersects the
x-axis at 12
, 0 , so the y-intercept is
(0, −1) and the x-intercept is 12
, 0 .
(b) An equation defining f is y = 2x − 1.
47. (a) The line falls 1 unit each time the x value increases by 3 units. Therefore the slope
is 13
. The graph intersects the y-axis at
the point (0, 2), so the y-intercept is (0, 2). The graph passes through (3, 1) and will fall 1 unit when the x value increases by 3, so the x-intercept is (6, 0).
(b) An equation defining f is 13
2.y x
48. (a) The line rises 3 units each time the x value increases by 4 units. Therefore the
slope is 34
. The graph intersects the
y-axis at the point (0, −3) and intersects the x-axis at (4, 0), so the y-intercept is (0, −3) and the x-intercept is 4.
(b) An equation defining f is 34
3.y x
49. (a) The line falls 200 units each time the x value increases by 1 unit. Therefore the slope is −200. The graph intersects the y-axis at the point (0, 300) and intersects
the x-axis at 32
, 0 , so the y-intercept is
(0, 300) and the x-intercept is 32
, 0 .
(b) An equation defining f is y = −200x + 300.
50. (a) The line rises 100 units each time the x value increases by 5 units. Therefore the slope is 20. The graph intersects the y-axis at the point (0, −50) and intersects
the x-axis at 52
, 0 , so the y-intercept is
(0, −50) and the x-intercept is 52
, 0 .
(b) An equation defining f is y = 20x − 50.
208 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
51. (a) through (–1, 4), parallel to x + 3y = 5 Find the slope of the line x + 3y = 5 by writing this equation in slope-intercept form.
513 3
3 5 3 – 5–
x y y xy x
The slope is 13
.
Because the lines are parallel, 13
is also
the slope of the line whose equation is to
be found. Substitute 13
m , 1 –1,x
and 1 4y into the point-slope form.
1 11313
– –
4 – –1
– 4 13 –12 – –1 3 11
y y m x xy xy x
y x x y
(b) Solve for y. 1 113 3
3 – 11 –y x y x
52. (a) through (3, –2), parallel to 2x – y = 5 Find the slope of the line 2x – y = 5 by writing this equation in slope-intercept form. 2 – 5 – –2 5
2 – 5x y y x
y x
The slope is 2. Because the lines are parallel, the slope of the line whose equation is to be found is also 2. Substitute m = 2, 1 3x , and 1 2y into the point-slope form.
1 1 – –
2 2 – 3 2 2 – 6–2 –8 or 2 – 8
y y m x xy x y x
x y x y
(b) Solve for y. y = 2x – 8
53. (a) through (1, 6), perpendicular to 3x + 5y = 1 Find the slope of the line 3x + 5y = 1 by writing this equation in slope-intercept form.
3 15 5
3 5 1 5 –3 1–
x y y xy x
This line has a slope of 35
. The slope of
any line perpendicular to this line is 53
,
because 3 55 3
1. Substitute 53
m ,
1 1,x and 1 6y into the point-slope form.
53
– 6 ( –1)3( – 6) 5( –1)3 – 18 5 – 5
–13 5 – 3 or 5 – 3 –13
y xy xy x
x y x y
(b) Solve for y. 5 133 3
3 5 13y x y x
54. (a) through (–2, 0), perpendicular to 8x – 3y = 7 Find the slope of the line 8x – 3y = 7 by writing the equation in slope-intercept form.
8 73 3
8 – 3 7 –3 –8 7–
x y y xy x
This line has a slope of 83
. The slope of
any line perpendicular to this line is 38
,
because 8 33 8
1.
Substitute 38
m , 1 2,x and 1 0y
into the point-slope form. 38
– 0 – ( 2)8 –3( 2)8 –3 – 6 3 8 –6
y xy xy x x y
(b) Solve for y. 3 68 8
3 38 4
8 –3 – 6 –
–
y x y xy x
55. (a) through (4, 1), parallel to y = −5 Because y = −5 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, 1), the equation is y = 1.
(b) The slope-intercept form is y = 1.
56. (a) through 2, 2 , parallel to y = 3.
Because y = 3 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through 2, 2 ,
the equation is y = –2.
(b) The slope-intercept form is y = −2
57. (a) through (–5, 6), perpendicular to x = –2. Because x = –2 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (–5, 6), the equation is y = 6.
Section 2.5 Equations of Lines and Linear Models 209
Copyright © 2017 Pearson Education, Inc.
(b) The slope-intercept form is y = 6.
58. (a) Through (4, –4), perpendicular to x = 4 Because x = 4 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, –4), the equation is y = –4.
(b) The slope-intercept form is y = –4.
59. (a) Find the slope of the line 3y + 2x = 6.
23
3 2 6 3 –2 6– 2
y x y xy x
Thus, 23
– .m A line parallel to
3y + 2x = 6 also has slope 23
– .
Solve for k using the slope formula.
12
2 1 2–
4 33 2
–4 3
3 23 4 3 4 –
4 39 2 49 2 8
2 1
k
k
k kk
kk
k k
(b) Find the slope of the line 2y – 5x = 1.
5 12 2
2 5 1 2 5 1y x y xy x
Thus, 52
.m A line perpendicular to 2y
– 5x = 1 will have slope 25
– , because
5 22 5
– –1.
Solve this equation for k.
72
3 2
4 53 2
5 4 5 44 515 2 415 2 82 7
k
k kk
kk
k k
60. (a) Find the slope of the line 2 – 3 4.x y
2 43 3
2 – 3 4 –3 –2 4–
x y y xy x
Thus, 23
.m A line parallel to
2x – 3y = 4 also has slope 23
. Solve for r
using the slope formula. – 6 2 – 6 2
– 4 – 2 3 – 6 3
– 6 26 6
– 6 36 4 2
r r
r
r r
(b) Find the slope of the line x + 2y = 1.
1 12 2
2 1 2 – 1–
x y y xy x
Thus, 12
.m A line perpendicular to
the line x + 2y = 1 has slope 2, because 12
– (2) –1. Solve for r using the slope
formula. – 6 – 6
2 2– 4 – 2 – 6
– 6 –12 – 6
r r
r r
61. (a) First find the slope using the points (0, 6312) and (3, 7703).
7703 6312 1391463.67
3 0 3m
The y-intercept is (0, 6312), so the equation of the line is
463.67 6312.y x
(b) The value x = 4 corresponds to the year 2013.
463.67 4 6312 8166.68y
The model predicts that average tuition and fees were $8166.68 in 2013. This is $96.68 more than the actual amount.
62. (a) First find the slope using the points (0, 6312) and (2, 7136).
7136 6312 824412
2 0 2m
The y-intercept is (0, 6312), so the equation of the line is 412 6312.y x
(b) The value x = 4 corresponds to the year 2013.
412 4 6312 7960y
The model predicts that average tuition and fees were $7960 in 2013. This is $110 less than the actual amount.
210 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
63. (a) First find the slope using the points (0, 22036) and (4, 24525).
24525 22036 2489622.25
4 0 4m
The y-intercept is (0, 22036), so the equation of the line is
622.25 22,036.y x
The slope of the line indicates that the
average tuition increase is about $622 per year from 2009 through 2013.
(b) The year 2012 corresponds to x = 3. 622.25 3 22,036 23,902.75y
According to the model, average tuition and fees were $23,903 in 2012. This is $443 more than the actual amount $23,460.
(c) Using the linear regression feature, the equation of the line of best fit is
653 21,634.y x
64. (a) See the graph in the answer to part (b).There appears to be a linear relationship between the data. The farther the galaxy is from Earth, the faster it is receding.
(b) Using the points (520, 40,000) and (0, 0), we obtain
40,000 – 0 40,00076.9.
520 – 0 520m
The equation of the line through these two points is y = 76.9x.
(c) 76.9 60,00060,000
78076.9
x
x x
According to the model, the galaxy Hydra is approximately 780 megaparsecs away.
(d) 11
1110 9
9.5 10
9.5 101.235 10 12.35 10
76.9
Am
A
Using m = 76.9, we estimate that the age of the universe is approximately 12.35 billion years.
(e) 11
10 9
119
9.5 101.9 10 or 19 10
509.5 10
9.5 10100
A
A
The range for the age of the universe is between 9.5 billion and 19 billion years.
65. (a) The ordered pairs are (0, 32) and (100, 212).
The slope is 212 – 32 180 9
.100 – 0 100 5
m
Use 91 1 5
( , ) (0,32) and x y m in the
point-slope form.
1 195959 95 5
– ( – )
– 32 ( – 0)
– 32
32 32
y y m x xy xy x
y x F C
(b)
95
59
32
5 9 325 9 160 9 5 –1609 5( – 32) ( – 32)
F CF CF C C FC F C F
(c) 59
( – 32)9 5( – 32) 9 5 – 1604 –160 – 40
F C F FF F F FF F
F = C when F is –40º.
Section 2.5 Equations of Lines and Linear Models 211
Copyright © 2017 Pearson Education, Inc.
66. (a) The ordered pairs are (0, 1) and (100, 3.92). The slope is
3.92 – 1 2.920.0292
100 – 0 100m and 1.b
Using slope-intercept form we have 0.0292 1y x or ( ) 0.0292 1.p x x
(b) Let x = 60. P(60) = 0.0292(60) + 1 = 2.752 The pressure at 60 feet is approximately 2.75 atmospheres.
67. (a) Because we want to find C as a function of I, use the points (12026, 10089) and (14167, 11484), where the first component represents the independent variable, I. First find the slope of the line.
11484 10089 13950.6516
14167 12026 2141m
Now use either point, say (12026, 10089), and the point-slope form to find the equation.
–10089 0.6516( –12026)–10089 0.6516 – 7836
0.6516 2253
C IC I
C I
(b) Because the slope is 0.6516, the marginal propensity to consume is 0.6516.
68. D is the only possible answer, because the x-intercept occurs when y = 0. We can see from the graph that the value of the x-intercept exceeds 10.
69. Write the equation as an equivalent equation with 0 on one side: 2 7 4 2x x x 2 7 4 2 0x x x . Now graph
2 7 4 2y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:
Solution set: {3}
70. Write the equation as an equivalent equation with 0 on one side: 7 2 4 5 3 1x x x 7 2 4 5 3 1 0x x x . Now graph
7 2 4 5 3 1y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:
Solution set: {1}
71. Write the equation as an equivalent equation with 0 on one side: 3 2 1 2 2 5x x
3 2 1 2 2 5 0x x . Now graph
3 2 1 2 2 5y x x in the window
[–5, 5] × [–5, 5] to find the x-intercept:
Solution set: 12
or 0.5
72. Write the equation as an equivalent equation with 0 on one side:
4 3 4 2 2 3 6 2x x x x
4 3 4 2 2 3 6 2 0x x x x .
Now graph 4 3 4 2 2 3 6 2y x x x x in the
window [–2, 8] × [–5, 5] to find the x-intercept:
Solution set: {4}
73. (a) 2 5 22 10 2
10 212
x xx x
xx
Solution set: {12}
212 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b) Answers will vary. Sample answer: The solution does not appear in the standard viewing window x-interval [10, –10]. The minimum and maximum values must include 12.
74. Rewrite the equation as an equivalent equation with 0 on one side.
3 2 6 4 8 2
6 18 4 8 2 0
x x xx x x
Now graph y = −6x − 18 − (−4x + 8 −2x) in the window [–10, 10] × [–30, 10].
The graph is a horizontal line that does not
intersect the x-axis. Therefore, the solution set is . We can verify this algebraically.
3 2 6 4 8 26 18 6 8 0 26
x x xx x
Because this is a false statement, the solution set is .
75. A(–1, 4), B(–2, –1), C(1, 14)
For A and B, 1 4 5
52 ( 1) 1
m
For B and C, 14 ( 1) 15
51 ( 2) 3
m
For A and C, 14 4 10
51 ( 1) 2
m
Since all three slopes are the same, the points are collinear.
76. A(0, −7), B(–3, 5), C(2, −15)
For A and B, 5 ( 7) 12
43 0 3
m
For B and C, 15 5 20
42 ( 3) 5
m
For A and C, 15 ( 7) 8
42 0 2
m
Since all three slopes are the same, the points are collinear.
77. A(−1, −3), B(−5, 12), C(1, −11)
For A and B, 12 ( 3) 15
5 ( 1) 4m
For B and C, 11 12 23
1 ( 5) 6m
For A and C, 11 ( 3) 8
41 ( 1) 2
m
Since all three slopes are not the same, the points are not collinear.
78. A(0, 9), B(–3, –7), C(2, 19)
For A and B, 7 9 16 16
3 0 3 3m
For B and C, 19 ( 7) 26
2 ( 3) 5m
For A and C, 19 9 10
52 0 2
m
Because all three slopes are not the same, the points are not collinear.
79. 2 21 1 1
2 2 21 1 1
( , ) ( – 0) ( – 0)d O P x m x
x m x
80. 2 22 2 2
2 2 22 2 2
( , ) ( – 0) ( – 0)d O Q x m x
x m x
81. 2 22 1 2 2 1 1( , ) ( – ) ( – )d P Q x x m x m x
82.
2 2 2
2 22 2 2 2 2 2
1 1 1 2 2 2
22 2
2 1 2 2 1 1
2 2 2 2 2 21 1 1 2 2 2
2 22 1 2 2 1 1
[ ( , )] [ ( , )] [ ( , )]
– –
– –
d O P d O Q d P Q
x m x x m x
x x m x m x
x m x x m x
x x m x m x
2 2 2 2 2 21 1 1 2 2 2
2 2 2 22 2 1 1 2 2
2 21 2 1 2 1 1
2 1 1 2 1 2
1 2 1 2 1
2
20 2 2
2 2 0
x m x x m xx x x x m x
m m x x m xx x m m x x
m m x x x x
83. 1 2 1 2 1 2
1 2 1 2
–2 – 2 0–2 ( 1) 0m m x x x x
x x m m
84. 1 2 1 2–2 ( 1) 0x x m m
Because 1 20 and 0, x x we have
1 2 1 21 0 implying that –1.m m m m
Summary Exercises on Graphs, Circles, Functions, and Equations 213
Copyright © 2017 Pearson Education, Inc.
85. If two nonvertical lines are perpendicular, then the product of the slopes of these lines is –1.
Summary Exercises on Graphs, Circles, Functions, and Equations
1. (3, 5), (2, 3)P Q
(a)
2 2
2 2
( , ) (2 3) ( 3 5)
1 8
1 64 65
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
5 33 2 5 2 5, , ,1
2 2 2 2 2
.
(c) First find m: 3 – 5 8
82 – 3 1
m
Use either point and the point-slope form. – 5 8 – 3y x
Change to slope-intercept form. 5 8 24 8 19y x y x
2. P(–1, 0), Q(4, –2)
(a) 2 2
2 2
( , ) [4 – (–1)] (–2 – 0)
5 (–2)
25 4 29
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
–1 4 0 (–2) 3 2, ,
2 2 2 23
, 1 .2
(c) First find m: 2 – 0 2 2
4 – 1 5 5m
Use either point and the point-slope form.
25
– 0 – 1y x
Change to slope-intercept form.
2 25 5
5 2 15 2 2
y xy xy x
3. P(–2, 2), Q(3, 2)
(a) 22
2 2
( , ) [3 – (– 2)] 2 2
5 0 25 0 25 5
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 2 3 2 2,
2 2
1 4 1, , 2 .
2 2 2
(c) First find m: 2 – 2 0
03 – 2 5
m
All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes through (3, 2), the equation is y = 2.
4. 2 2, 2 ,P 2,3 2Q
(a)
2 2
2 2
( , ) 2 – 2 2 3 2 – 2
– 2 2 2
2 8 10
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
2 2 2 2 3 2,
2 2
3 2 4 2 3 2, , 2 2 .
2 2 2
(c) First find m: 3 2 2 2 2
22 2 2 2
m
Use either point and the point-slope form.
– 2 2 – 2 2y x
Change to slope-intercept form.
– 2 2 4 2 2 5 2y x y x
5. P(5, –1), Q(5, 1)
(a) 2 2
2 2
( , ) (5 – 5) [1 – (–1)]
0 2 0 4 4 2
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
5 5 –1 1,
2 2
10 0
, 5,0 .2 2
214 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(c) First find m. 1 – 1 2
undefined5 – 5 0
m
All lines that have an undefined slope are vertical lines. The equation of a vertical line has an equation of the form x = a. The line passes through (5, 1), so the equation is x = 5 . (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)
6. (1, 1), ( 3, 3)P Q
(a)
2 2
2 2
( , ) ( 3 – 1) ( 3 –1)
4 4
16 16 32 4 2
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
1 3 1 3,
2 2
2 2,
2 2
1, 1 .
(c) First find m: 3 –1 4
13 –1 4
m
Use either point and the point-slope form. –1 1 –1y x
Change to slope-intercept form. 1 1y x y x
7. 2 3,3 5 , 6 3,3 5P Q
(a)
2 2
2 2
( , ) 6 3 – 2 3 3 5 – 3 5
4 3 0 48 4 3
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
2 3 6 3 3 5 3 5,
2 2
8 3 6 5, 4 3, 3 5 .
2 2
(c) First find m: 3 5 3 5 0
06 3 2 3 4 3
m
All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes
through 2 3,3 5 , the equation is
3 5.y
8. P(0, –4), Q(3, 1)
(a) 2 2
2 2
( , ) (3 – 0) [1 – (– 4)]
3 5 9 25 34
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
0 3 – 4 1 3 3 3 3, , , .
2 2 2 2 2 2
(c) First find m: 1 – 4 5
3 0 3m
Using slope-intercept form we have 53
– 4.y x
9. Through 2,1 and 4, 1
First find m: 1 –1 2 1
4 – (–2) 6 3m
Use either point and the point-slope form.
13
– 1 – 4y x
Change to slope-intercept form.
1 13 3
3 1 4 3 3 4
3 1
y x y xy x y x
Summary Exercises on Graphs, Circles, Functions, and Equations 215
Copyright © 2017 Pearson Education, Inc.
10. the horizontal line through (2, 3) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (2, 3), the equation is y = 3.
11. the circle with center (2, –1) and radius 3
22 2
2 2
( – 2) – (–1) 3
( – 2) ( 1) 9
x yx y
12. the circle with center (0, 2) and tangent to the x-axis The distance from the center of the circle to the x-axis is 2, so r = 2.
2 2 2 2 2( – 0) ( – 2) 2 ( – 2) 4x y x y
13. the line through (3, 5) with slope 56
Write the equation in point-slope form.
56
– 5 3 y x
Change to standard form.
5 156 6
5 56 2
6 5 5 3 6 30 5 15
6 5 15
y x y xy x y x
y x
14. the vertical line through (−4, 3) The equation of a vertical line has an equation of the form x = a. Because the line passes through (−4, 3), the equation is x = −4.
15. a line through (–3, 2) and parallel to the line 2x + 3y = 6 First, find the slope of the line 2x + 3y = 6 by writing this equation in slope-intercept form.
23
2 3 6 3 2 6 2x y y x y x
The slope is 23
. Because the lines are
parallel, 23
is also the slope of the line
whose equation is to be found. Substitute 23
m , 1 3x , and 1 2y into the point-
slope form.
21 1 3
23
– – 2 3
3 2 2 3 3 6 2 – 6
3 –2
y y m x x y xy x y x
y x y x
16. a line through the origin and perpendicular to the line 3 4 2x y
First, find the slope of the line 3 4 2x y by writing this equation in slope-intercept form.
3 32 14 4 4 2
3 4 2 4 –3 2x y y xy x y x
This line has a slope of 34
. The slope of any
line perpendicular to this line is 43
, because
343 4
1. Using slope-intercept form we
have 4 43 3
0 or .y x y x
(continued on next page)
216 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
17. 2 24 2 4x x y y Complete the square on x and y separately.
2 2
2 2
2 2
4 2 4
4 4 2 1 4 4 1
2 1 9
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (2, −1) and radius 3.
18. 2 26 10 36 0x x y y Complete the square on x and y separately.
2 2
2 2
2 2
6 10 36
6 9 10 25 –36 9 25
3 5 2
x x y y
x x y y
x y
No, it is not a circle.
19. 2 212 20 0x x y Complete the square on x and y separately.
2 2
2 2
2 2
12 20
12 36 –20 36
6 16
x x y
x x y
x y
Yes, it is a circle. The circle has its center at (6, 0) and radius 4.
20. 2 22 16 61x x y y Complete the square on x and y separately.
2 2
2 2
2 2
2 16 61
2 1 16 64 –61 1 64
1 8 4
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (−1, −8) and radius 2.
21. 2 22 10 0x x y Complete the square on x and y separately.
2 2
2 2
2 2
2 10
2 1 –10 1
1 9
x x y
x x y
x y
No, it is not a circle.
22. 2 2 – 8 9 0x y y Complete the square on x and y separately.
2 2
2 2
22
– 8 9
– 8 16 9 16
– 4 25
x y y
x y y
x y
Yes, it is a circle. The circle has its center at (0, 4) and radius 5.
23. The equation of the circle is 2 2 2( 4) ( 5) 4 .x y
Let y = 2 and solve for x: 2 2 2( 4) (2 5) 4x 2 2 2 2( 4) ( 3) 4 ( 4) 7x x
4 7 4 7x x
The points of intersection are 4 7, 2 and
4 7, 2
24. Write the equation in center-radius form by completing the square on x and y separately:
2 2
2 2
2 2
2 2
10 24 144 0
10 24 144 0
( 10 25) ( 24 144) 25
( 5) ( 12) 25
x y x yx x y y
x x y yx y
The center of the circle is (5, 12) and the radius is 5. Now use the distance formula to find the distance from the center (5, 12) to the origin:
2 22 1 2 1
2 2(5 0) (12 0) 25 144 13
d x x y y
The radius is 5, so the shortest distance from the origin to the graph of the circle is 13 − 5 = 8.
(continued on next page)
Section 2.6 Graphs of Basic Functions 217
Copyright © 2017 Pearson Education, Inc.
(continued)
25. (a) The equation can be rewritten as 6 31 1
4 4 4 24 6 .y x y x y x
x can be any real number, so the domain is all real numbers and the range is also all real numbers. domain: , ; range: ,
(b) Each value of x corresponds to just one value of y. 4 6x y represents a function.
3 31 14 2 4 2
y x f x x
3 31 1 24 2 2 2 2
2 2 1f
26. (a) The equation can be rewritten as 2 5 .y x y can be any real number.
Because the square of any real number is
not negative, 2y is never negative. Taking the constant term into consideration, domain would be 5, .
domain: 5, ; range: ,
(b) Because (−4, 1) and (−4, −1) both satisfy
the relation, 2 5y x does not represent a function.
27. (a) 2 22 25x y is a circle centered at
(−2, 0) with a radius of 5. The domain will start 5 units to the left of –2 and end 5 units to the right of –2. The domain will be [−2 − 5, 2 + 5] = [−7, 3]. The range will start 5 units below 0 and end 5 units above 0. The range will be [0 − 5, 0 + 5] = [−5, 5].
(b) Because (−2, 5) and (−2, −5) both satisfy
the relation, 2 22 25x y does not
represent a function.
28. (a) The equation can be rewritten as 2 2 31
2 22 3 .y x y x x can be
any real number. Because the square of
any real number is not negative, 212
x is
never negative. Taking the constant term into consideration, range would be
32
, .
domain: , ; range: 32
,
(b) Each value of x corresponds to just one
value of y. 2 2 3x y represents a function.
2 23 31 12 2 2 2
y x f x x
2 3 3 31 1 4 12 2 2 2 2 2 2
2 2 4f
Section 2.6 Graphs of Basic Functions
1. The equation 2f x x matches graph E.
The domain is , .
2. The equation of f x x matches graph G.
The function is increasing on 0, .
3. The equation 3f x x matches graph A.
The range is , .
4. Graph C is not the graph of a function.
Its equation is 2x y .
5. Graph F is the graph of the identity function. Its equation is .f x x
6. The equation f x x matches graph B.
1.5 1f
7. The equation 3f x x matches graph H.
No, there is no interval over which the function is decreasing.
8. The equation of f x x matches graph D.
The domain is 0, .
9. The graph in B is discontinuous at many points. Assuming the graph continues, the range would be {…, −3, −2, −1, 0, 1, 2, 3, …}.
218 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
10. The graphs in E and G decrease over part of the domain and increase over part of the domain. They both increase over 0, and
decrease over , 0 .
11. The function is continuous over the entire domain of real numbers , .
12. The function is continuous over the entire domain of real numbers , .
13. The function is continuous over the interval
0, .
14. The function is continuous over the interval , 0 .
15. The function has a point of discontinuity at (3, 1). It is continuous over the interval ,3 and the interval 3, .
16. The function has a point of discontinuity at x = 1. It is continuous over the interval ,1
and the interval 1, .
17. 2 if –1( )
–1 if –1x xf x x x
(a) (–5) 2(–5) –10f
(b) (–1) 2(–1) –2f
(c) f(0) = 0 –1 = –1
(d) f(3) = 3 – 1 = 2
18. – 2 if 3( )
5 – if 3x xf x x x
(a) f(–5) = –5 – 2 = –7
(b) f(–1) = –1 – 2 = –3
(c) f(0) = 0 – 2 = –2
(d) f(3) = 5 – 3 = 2
19. 2 if – 4
( ) – if – 4 23 if 2
x xf x x x
x x
(a) f(–5) = 2 + (–5) = –3
(b) f(–1) = – (–1) = 1
(c) f(0) = –0 = 0
(d) (3) 3 3 9f
20. –2 if –33 –1 if – 3 2– 4 if 2
x xf x x x
x x
(a) f(–5) = –2(–5) = 10
(b) f(–1) = 3(–1) – 1 = –3 – 1 = –4
(c) f(0) = 3(0) – 1 = 0 – 1 = –1
(d) f(3) = –4(3) = –12
21. – 1 if 3( )
2 if 3x xf x x
Draw the graph of y = x – 1 to the left of x = 3, including the endpoint at x = 3. Draw the graph of y = 2 to the right of x = 3, and note that the endpoint at x = 3 coincides with the endpoint of the other ray.
22. 6 – if 3( )
3 if 3x xf x x
Graph the line y = 6 – x to the left of x = 3, including the endpoint. Draw y = 3 to the right of x = 3. Note that the endpoint at x = 3 coincides with the endpoint of the other ray.
23. 4 – if 2( )
1 2 if 2x xf x x x
Draw the graph of y = 4 − x to the left of x = 2, but do not include the endpoint. Draw the graph of y = 1 + 2x to the right of x = 2, including the endpoint.
Section 2.6 Graphs of Basic Functions 219
Copyright © 2017 Pearson Education, Inc.
24. 2 1 if 0( )
if 0x xf x x x
Graph the line y = 2x + 1 to the right of x = 0, including the endpoint. Draw y = x to the left of x = 0, but do not include the endpoint.
25. 3 if 1( )
1 if 1xf x x
Graph the line y = −3 to the left of x = 1, including the endpoint. Draw y = −1 to the right of x = 1, but do not include the endpoint.
26. 2 if 1( )
2 if 1xf x x
Graph the line y = – 2 to the left of x = 1, including the endpoint. Draw y = 2 to the right of x = 1, but do not include the endpoint.
Exercise 26 Exercise 27
27. 2 if – 4
( ) – if – 4 53 if 5
x xf x x x
x x
Draw the graph of y = 2 + x to the left of –4, but do not include the endpoint at x = 4. Draw the graph of y = –x between –4 and 5, including both endpoints. Draw the graph of y = 3x to the right of 5, but do not include the endpoint at x = 5.
28. –2 if –3
( ) 3 –1 if – 3 2– 4 if 2
x xf x x x
x x
Graph the line y = –2x to the left of x = –3, but do not include the endpoint. Draw y = 3x – 1 between x = –3 and x = 2, and include both endpoints. Draw y = –4x to the right of x = 2, but do not include the endpoint. Notice that the endpoints of the pieces do not coincide.
29. 21
212
– +2 if 2( )
if 2
x xf x
x x
Graph the curve 212
2y x to the left of
x = 2, including the endpoint at (2, 0). Graph
the line 12
y x to the right of x = 2, but do
not include the endpoint at (2, 1). Notice that the endpoints of the pieces do not coincide.
30. 3
25 if 0
( )if 0
x xf xx x
Graph the curve 3 5y x to the left of x = 0, including the endpoint at (0, 5). Graph
the line 2y x to the right of x = 0, but do not include the endpoint at (0, 0). Notice that the endpoints of the pieces do not coincide.
220 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
31. 2
2 if 5 1( ) 2 if 1 0
2 if 0 2
x xf x x
x x
Graph the line 2y x between x = −5 and x = −1, including the left endpoint at (−5, −10), but not including the right endpoint at (−1, −2). Graph the line y = −2 between x = −1 and x = 0, including the left endpoint at (−1, −2) and not including the right endpoint at (0, −2). Note that (−1, −2) coincides with the first two sections, so it is included. Graph
the curve 2 2y x from x = 0 to x = 2, including the endpoints at (0, −2) and (2, 2).
Note that (0, −2) coincides with the second two sections, so it is included. The graph ends at x = −5 and x = 2.
32.
2
2
0.5 if 4 2( ) if 2 2
4 if 2 4
x xf x x x
x x
Graph the curve 20.5y x between x = −4 and x = −2, including the endpoints at (−4, 8) and (−2, 2). Graph the line y x between x = −2 and x = 2, but do not include the endpoints at (−2, −2) and (2, 2). Graph the
curve 2 4y x from x = 2 to x = 4, including the endpoints at (2, 0) and (4, 12). The graph ends at x = −4 and x = 4.
33.
3
2
3 if 2 0( ) 3 if 0 1
4 if 1 3
x xf x x x
x x x
Graph the curve 3 3y x between x = −2 and x = 0, including the endpoints at (−2, −5) and (0, 3). Graph the line y = x + 3 between x = 0 and x = 1, but do not include the endpoints at (0, 3) and (1, 4). Graph the curve
24y x x from x = 1 to x = 3, including the endpoints at (1, 4) and (3, −2). The graph ends at x = −2 and x = 3.
34. 2
312
2 if 3 1
( ) 1 if 1 2
1 if 2 3
x xf x x x
x x
Graph the curve y = −2x to from x = −3 to x = −1, including the endpoint (−3, 6), but not including the endpoint (−1, 2). Graph the
curve 2 1y x from x = −1 to x = 2, including the endpoints (−1, 2) and (2, 5).
Graph the curve 312
1y x from x = 2 to
x = 3, including the endpoint (3, 14.5) but not including the endpoint (2, 5). Because the endpoints that are not included coincide with endpoints that are included, we use closed dots on the graph.
Section 2.6 Graphs of Basic Functions 221
Copyright © 2017 Pearson Education, Inc.
35. The solid circle on the graph shows that the endpoint (0, –1) is part of the graph, while the open circle shows that the endpoint (0, 1) is not part of the graph. The graph is made up of parts of two horizontal lines. The function which fits this graph is
–1 if 0( )
1 if 0.xf x x
domain: , ; range: {–1, 1}
36. We see that y = 1 for every value of x except x = 0, and that when x = 0, y = 0. We can write the function as
1 if 0( )
0 if 0.xf x x
domain: , ; range: {0, 1}
37. The graph is made up of parts of two horizontal lines. The solid circle shows that the endpoint (0, 2) of the one on the left belongs to the graph, while the open circle shows that the endpoint (0, –1) of the one on the right does not belong to the graph. The function that fits this graph is
2 if 0( )
–1 if 1.xf x x
domain: ( , 0] (1, ); range: {–1, 2}
38. We see that y = 1 when 1x and that y = –1 when x > 2. We can write the function as
1 if –1( )
–1 if 2.xf x x
domain: (– , –1] (2, ); range: {–1, 1}
39. For 0x , that piece of the graph goes through the points (−1, −1) and (0, 0). The slope is 1, so the equation of this piece is y = x. For x > 0, that piece of the graph is a horizontal line passing through (2, 2), so its equation is y = 2. We can write the function as
if 0( ) .
2 if 0x xf x x
domain: (– , ) range: ( , 0] {2}
40. For x < 0, that piece of the graph is a horizontal line passing though (−3, −3), so the equation of this piece is y = −3. For 0x , the curve passes through (1, 1) and (4, 2), so the
equation of this piece is y x . We can
write the function as 3 if 0
( ) . if 0
xf x
x x
domain: (– , ) range: { 3} [0, )
41. For x < 1, that piece of the graph is a curve passes through (−8, −2), (−1, −1) and (1, 1), so
the equation of this piece is 3y x . The right piece of the graph passes through (1, 2) and
(2, 3). 2 3
11 2
m
, and the equation of the
line is 2 1 1y x y x . We can write
the function as 3 if 1( )
1 if 1x xf x
x x
domain: (– , ) range: ( ,1) [2, )
42. For all values except x = 2, the graph is a line. It passes through (0, −3) and (1, −1). The slope is 2, so the equation is y = 2x −3. At x = 2, the graph is the point (2, 3). We can write
the function as 3 if 2( ) .
2 3 if 2xf x x x
domain: (– , ) range: ( ,1) (1, )
43. f(x) = x
Plot points.
x –x f(x) = x
–2 2 2
–1.5 1.5 1
–1 1 1
–0.5 0.5 0
0 0 0
0.5 –0.5 –1
1 –1 –1
1.5 –1.5 –2
2 –2 –2
More generally, to get y = 0, we need 0 – 1 0 1 1 0.x x x To get y = 1, we need 1 – 2x
1 2 –2 1.x x Follow this pattern to graph the step function.
domain: , ; range: {...,–2,–1,0,1,2,...}
222 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
44. f(x) = x
Plot points.
x x f(x) = x
–2 −2 2
–1.5 −2 2
–1 −1 1
–0.5 −1 1
0 0 0
0.5 0 0
1 1 −1
1.5 1 –1
2 2 –2
Follow this pattern to graph the step function.
domain: , ; range: {...,–2,–1,0,1,2,...}
45. f(x) = 2x
To get y = 0, we need 12
0 2 1 0 .x x
To get y = 1, we need 12
1 2 2 1.x x
To get y = 2, we need 32
2 2 3 1 .x x
Follow this pattern to graph the step function.
domain: , ; range: {...,–2,–1,0,1,2,...}
46. g(x) = 2 1x
To get y = 0, we need 12
0 2 1 1 1 2 2 1.x x x
To get y = 1, we need 32
1 2 – 1 2 2 2 3 1 .x x x
Follow this pattern to graph the step function.
domain: , ; range: {..., 2,–1,0,1,2,...}
47. The cost of mailing a letter that weighs more than 1 ounce and less than 2 ounces is the same as the cost of a 2-ounce letter, and the cost of mailing a letter that weighs more than 2 ounces and less than 3 ounces is the same as the cost of a 3-ounce letter, etc.
48. The cost is the same for all cars parking
between 12
hour and 1-hour, between 1 hour
and 12
1 hours, etc.
49.
Section 2.7 Graphing Techniques 223
Copyright © 2017 Pearson Education, Inc.
50.
51. (a) For 0 8,x 49.8 34.2
1.958 0
m
,
so 1.95 34.2.y x For 8 13x ,
52.2 49.80.48
13 8m
, so the equation
is 52.2 0.48( 13)y x 0.48 45.96y x
(b) 1.95 34.2 if 0 8( )
0.48 45.96 if 8 13x xf x x x
52. When 0 3x , the slope is 5, which means that the inlet pipe is open, and the outlet pipe is closed. When 3 5x , the slope is 2, which means that both pipes are open. When 5 8x , the slope is 0, which means that both pipes are closed. When 8 10x , the slope is −3, which means that the inlet pipe is closed, and the outlet pipe is open.
53. (a) The initial amount is 50,000 gallons. The final amount is 30,000 gallons.
(b) The amount of water in the pool remained constant during the first and fourth days.
(c) (2) 45,000; (4) 40,000f f
(d) The slope of the segment between (1, 50000) and (3, 40000) is −5000, so the water was being drained at 5000 gallons per day.
54. (a) There were 20 gallons of gas in the tank at x = 3.
(b) The slope is steepest between t = 1 and t ≈ 2.9, so that is when the car burned gasoline at the fastest rate.
55. (a) There is no charge for additional length, so we use the greatest integer function. The cost is based on multiples of two
feet, so 2
( ) 0.8 xf x if 6 18x .
(b) 8.52
(8.5) 0.8 0.8(4) $3.20f
15.22
(15.2) 0.8 0.8(7) $5.60f
56. (a) 6.5 if 0 4
( ) 5.5 48 if 4 6–30 195 if 6 6.5
x xf x x x
x x
Draw a graph of y = 6.5x between 0 and 4, including the endpoints. Draw the graph of y = –5.5x + 48 between 4 and 6, including the endpoint at 6 but not the one at 4. Draw the graph of y = –30x + 195, including the endpoint at 6.5 but not the one at 6. Notice that the endpoints of the three pieces coincide.
(b) From the graph, observe that the snow depth, y, reaches its deepest level (26 in.) when x = 4, x = 4 represents 4 months after the beginning of October, which is the beginning of February.
(c) From the graph, the snow depth y is nonzero when x is between 0 and 6.5. Snow begins at the beginning of October and ends 6.5 months later, in the middle of April.
Section 2.7 Graphing Techniques
1. To graph the function 2 3,f x x shift the
graph of 2y x down 3 units.
2. To graph the function 2 5,f x x shift the
graph of 2y x up 5 units.
3. The graph of 24f x x is obtained by
shifting the graph of 2y x to the left 4 units.
4. The graph of 27f x x is obtained by
shifting the graph of 2y x to the right 7 units.
5. The graph of f x x is a reflection of
the graph of f x x across the x-axis.
6. The graph of f x x is a reflection of
the graph of f x x across the y-axis.
224 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
7. To obtain the graph of 32 3,f x x
shift the graph of 3y x 2 units to the left and 3 units down.
8. To obtain the graph of 33 6,f x x
shift the graph of 3y x 3 units to the right and 6 units up.
9. The graph of f x x is the same as the
graph of y x because reflecting it across
the y-axis yields the same ordered pairs.
10. The graph of 2x y is the same as the graph
of 2x y because reflecting it across the
x-axis yields the same ordered pairs.
11. (a) B; 2( 7)y x is a shift of 2 ,y x 7 units to the right.
(b) D; 2 7y x is a shift of 2 ,y x 7 units downward.
(c) E; 27y x is a vertical stretch of 2 ,y x by a factor of 7.
(d) A; 2( 7)y x is a shift of 2 ,y x 7 units to the left.
(e) C; 2 7y x is a shift of 2 ,y x 7 units upward.
12. (a) E; 34y x is a vertical stretch
of 3 ,y x by a factor of 4.
(b) C; 3y x is a reflection of 3 ,y x over the x-axis.
(c) D; 3y x is a reflection of 3 ,y x over the y-axis.
(d) A; 3 4y x is a shift of 3 ,y x 4 units to the right.
(e) B; 3 4y x is a shift of 3 ,y x 4 units down.
13. (a) B; 2 2y x is a shift of 2 ,y x 2 units upward.
(b) A; 2 2y x is a shift of 2 ,y x 2 units downward.
(c) G; 2( 2)y x is a shift of 2 ,y x 2 units to the left.
(d) C; 2( 2)y x is a shift of 2 ,y x 2 units to the right.
(e) F; 22y x is a vertical stretch of 2 ,y x by a factor of 2.
(f) D; 2y x is a reflection of 2 ,y x across the x-axis.
(g) H; 2( 2) 1y x is a shift of 2 ,y x 2 units to the right and 1 unit upward.
(h) E; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit upward.
(i) I; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit down.
14. (a) G; 3y x is a shift of ,y x 3 units to the left.
(b) D; 3y x is a shift of ,y x 3 units downward.
(c) E; 3y x is a shift of ,y x 3 units upward.
(d) B; 3y x is a vertical stretch of
,y x by a factor of 3.
(e) C; y x is a reflection of y x across the x-axis.
(f) A; 3y x is a shift of ,y x 3 units to the right.
(g) H; 3 2y x is a shift of ,y x 3 units to the right and 2 units upward.
(h) F; 3 2y x is a shift of ,y x 3 units to the left and 2 units upward.
(i) I; 3 2y x is a shift of ,y x 3 units to the right and 2 units downward.
15. (a) F; 2y x is a shift of y x 2 units
to the right.
(b) C; 2y x is a shift of y x 2 units
downward.
(c) H; 2y x is a shift of y x 2 units
upward.
Section 2.7 Graphing Techniques 225
Copyright © 2017 Pearson Education, Inc.
(d) D; 2y x is a vertical stretch of y x
by a factor of 2.
(e) G; y x is a reflection of
y x across the x-axis.
(f) A; y x is a reflection of y x
across the y-axis.
(g) E; 2y x is a reflection of 2y x
across the x-axis. 2y x is a vertical
stretch of y x by a factor of 2.
(h) I; 2 2y x is a shift of y x 2
units to the right and 2 units upward.
(i) B; 2 2y x is a shift of y x 2
units to the left and 2 units downward.
16. The graph of 32 1 6f x x is the graph
of 3f x x stretched vertically by a factor
of 2, shifted left 1 unit and down 6 units.
17. 3f x x
x h x x 3f x x
−2 2 6
−1 1 3
0 0 0
1 1 3
2 2 6
18. 4f x x
x h x x 4f x x
−2 2 8
−1 1 4
0 0 0
1 1 4
2 2 8
19. 23
f x x
x h x x 23
f x x
−3 3 2
−2 2 43
−1 1 23
0 0 0
1 1 23
2 2 43
3 3 2
20. 34
f x x
x h x x 34
f x x
−4 4 3
−3 3 94
−2 2 32
−1 1 34
0 0 0
1 1 34
2 2 32
3 3 94
4 4 3
(continued on next page)
226 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
21. 22f x x
x 2h x x 22f x x
−2 4 8
−1 1 2
0 0 0
1 1 2
2 4 8
22. 23f x x
x 2h x x 23f x x
−2 4 12
−1 1 3
0 0 0
1 1 3
2 4 12
23. 212
f x x
x 2h x x 212
f x x
−2 4 2
−1 1 12
0 0 0
1 1 12
2 4 2
24. 213
f x x
x 2h x x 213
f x x
−3 9 3
−2 4 43
−1 1 13
0 0 0
1 1 13
2 4 43
3 9 3
Section 2.7 Graphing Techniques 227
Copyright © 2017 Pearson Education, Inc.
25. 212
f x x
x 2h x x 212
f x x
−3 9 92
−2 4 2
−1 1 12
0 0 0
1 1 12
2 4 2
3 9 92
26. 213
f x x
x 2h x x 213
f x x
−3 9 −3
−2 4 43
−1 1 13
0 0 0
1 1 13
2 4 43
3 9 −3
27. 3f x x
x h x x 3f x x
−2 2 −6
−1 1 −3
0 0 0
1 1 −3
2 2 −6
28. 2f x x
x h x x 2f x x
−2 2 −4
−1 1 −2
0 0 0
1 1 −2
2 2 −4
29. 12
h x x
x f x x 1
21 12 2
h x x
x x
−4 4 2
−3 3 32
−2 2 1
−1 1 12
0 0 0
(continued on next page)
228 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
x f x x 1
21 12 2
h x x
x x
1 1 12
2 2 1
3 3 32
4 4 2
30. 13
h x x
x 13
f x x 1
31 13 3
h x x
x x
−3 3 1
−2 2 23
−1 1 13
0 0 0
1 1 13
2 2 23
3 3 1
31. 4h x x
x f x x 4 2h x x x
0 0 0
1 1 2
2 2 2 2
x f x x 4 2h x x x
3 3 2 3
4 2 4
32. 9h x x
x f x x 9 3h x x x
0 0 0
1 1 3
2 2 3 2
3 3 3 3
4 2 6
33. f x x
x h x x f x x
−4 2 −2
−3 3 3
−2 2 2
−1 1 −1
0 0 0
Section 2.7 Graphing Techniques 229
Copyright © 2017 Pearson Education, Inc.
34. f x x
x h x x f x x
−3 3 −3
−2 2 −2
−1 1 −1
0 0 0
1 1 −1
2 2 −2
3 3 −3
35. (a) 4y f x is a horizontal translation
of f, 4 units to the left. The point that corresponds to (8, 12) on this translated function would be 8 4,12 4,12 .
(b) 4y f x is a vertical translation of f, 4 units up. The point that corresponds to (8, 12) on this translated function would be 8,12 4 8,16 .
36. (a) 14
y f x is a vertical shrinking of f, by
a factor of 14
. The point that corresponds
to (8, 12) on this translated function
would be 14
8, 12 8,3 .
(b) 4y f x is a vertical stretching of f, by
a factor of 4. The point that corresponds to (8, 12) on this translated function would be 8, 4 12 8, 48 .
37. (a) (4 )y f x is a horizontal shrinking of f, by a factor of 4. The point that corresponds to (8, 12) on this translated
function is 14
8 , 12 2, 12 .
(b) 14
y f x is a horizontal stretching of f, by a factor of 4. The point that corresponds to (8, 12) on this translated function is 8 4, 12 32, 12 .
38. (a) The point that corresponds to (8, 12) when reflected across the x-axis would be (8, −12).
(b) The point that corresponds to (8, 12) when reflected across the y-axis would be (−8, 12).
39. (a) The point that is symmetric to (5, –3) with respect to the x-axis is (5, 3).
(b) The point that is symmetric to (5, –3) with respect to the y-axis is (–5, –3).
(c) The point that is symmetric to (5, –3) with respect to the origin is (–5, 3).
40. (a) The point that is symmetric to (–6, 1) with respect to the x-axis is (–6, –1).
(b) The point that is symmetric to (–6, 1) with respect to the y-axis is (6, 1).
(c) The point that is symmetric to (–6, 1) with respect to the origin is (6, –1).
41. (a) The point that is symmetric to (–4, –2) with respect to the x-axis is (–4, 2).
(b) The point that is symmetric to (–4, –2) with respect to the y-axis is (4, –2).
(c) The point that is symmetric to (–4, –2) with respect to the origin is (4, 2).
230 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
42. (a) The point that is symmetric to (–8, 0) with respect to the x-axis is (–8, 0) because this point lies on the x-axis.
(b) The point that is symmetric to the point (–8, 0) with respect to the y-axis is (8, 0).
(c) The point that is symmetric to the point (–8, 0) with respect to the origin is (8, 0).
43. The graph of y = |x − 2| is symmetric with respect to the line x = 2.
44. The graph of y = −|x + 1| is symmetric with respect to the line x = −1.
45. 2 5y x Replace x with –x to obtain
2 2( ) 5 5y x x . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
2 2 2( ) 2 2 2.y x y x y x The result is not the same as the original
equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.
46. 42 3y x Replace x with –x to obtain
4 42( ) 3 2 3y x x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to
obtain 4 4– 2( ) 3 2 3y x y x 42 3y x . The result is not the same as
the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.
47. 2 2 12x y Replace x with –x to obtain
2 2 2 2( ) 12 12x y x y . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2 2 2( ) 12 12x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin.
48. 2 2 6y x Replace x with –x to obtain
22 2 26 6y x y x
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2 2 2( ) 6 6y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis, the y-axis, and the origin.
49. 34y x x Replace x with –x to obtain
3 3
3
4( ) ( ) 4( )
4 .
y x x y x xy x x
The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to
obtain 3 34 4 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
3 3
3 34( ) ( ) 4( )
4 4 .
y x x y x xy x x y x x
The result is the same as the original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.
Section 2.7 Graphing Techniques 231
Copyright © 2017 Pearson Education, Inc.
50. 3y x x Replace x with –x to obtain
3 3( ) ( ) .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to
obtain 3 3 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to
obtain 3 3( ) ( )y x x y x x 3 .y x x The result is the same as the
original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.
51. 2 8y x x Replace x with –x to obtain
2 2( ) ( ) 8 8.y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y
to obtain 2( ) ( ) 8y x x 2 28 8.y x x y x x
The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
52. y = x + 15 Replace x with –x to obtain
( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain ( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
53. 3 2f x x x
3
3 3
2
2 2
f x x xx x x x f x
The function is odd.
54. 5 32f x x x
5 3
5 3 5 3
2
2 2
f x x xx x x x f x
The function is odd.
55. 4 20.5 2 6f x x x
4 2
4 2
0.5 2 6
0.5 2 6
f x x xx x f x
The function is even.
56. 20.75 4f x x x
2
2
0.75 4
0.75 4
f x x xx x f x
The function is even.
57. 3 9f x x x
3
3 3
9
9 9
f x x xx x x x f x
The function is neither.
58. 4 5 8f x x x
4
4
5 8
5 8
f x x xx x f x
The function is neither.
59. 2 1f x x
This graph may be obtained by translating the
graph of 2y x 1 unit downward.
60. 2 2f x x
This graph may be obtained by translating the
graph of 2y x 2 units downward.
232 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
61. 2 2f x x
This graph may be obtained by translating the
graph of 2y x 2 units upward.
62. 2 3f x x
This graph may be obtained by translating the
graph of 2y x 3 units upward.
63. 24g x x
This graph may be obtained by translating the
graph of 2y x 4 units to the right.
64. 22g x x
This graph may be obtained by translating the
graph of 2y x 2 units to the right.
65. 22g x x
This graph may be obtained by translating the
graph of 2y x 2 units to the left.
66. 23g x x
This graph may be obtained by translating the
graph of 2y x 3 units to the left.
67. 1g x x
The graph is obtained by translating the graph of y x 1 unit downward.
68. 3 2g x x
This graph may be obtained by translating the graph of y x 3 units to the left and 2 units
upward.
Section 2.7 Graphing Techniques 233
Copyright © 2017 Pearson Education, Inc.
69. 3( 1)h x x
This graph may be obtained by translating the
graph of 3y x 1 unit to the left. It is then reflected across the x-axis.
70. 3( 1)h x x
This graph can be obtained by translating the
graph of 3y x 1 unit to the right. It is then reflected across the x-axis. (We may also reflect the graph about the x-axis first and then translate it 1 unit to the right.)
71. 22 1h x x
This graph may be obtained by translating the
graph of 2y x 1 unit down. It is then stretched vertically by a factor of 2.
72. 23 2h x x
This graph may be obtained by stretching the
graph of 2y x vertically by a factor of 3, then shifting the resulting graph down 2 units.
73. 22( 2) 4f x x
This graph may be obtained by translating the
graph of 2y x 2 units to the right and 4 units down. It is then stretched vertically by a factor of 2.
74. 23( 2) 1f x x
This graph may be obtained by translating the
graph of 2y x 2 units to the right and 1 unit up. It is then stretched vertically by a factor of 3 and reflected over the x-axis.
75. 2f x x
This graph may be obtained by translating the
graph of y x two units to the left.
234 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
76. 3f x x
This graph may be obtained by translating the
graph of y x three units to the right.
77. f x x
This graph may be obtained by reflecting the
graph of y x across the x-axis.
78. 2f x x
This graph may be obtained by translating the
graph of y x two units down.
79. 2 1f x x
This graph may be obtained by stretching the
graph of y x vertically by a factor of two and then translating the resulting graph one unit up.
80. 3 2f x x
This graph may be obtained by stretching the
graph of y x vertically by a factor of three and then translating the resulting graph two units down.
81. 312
4g x x
This graph may be obtained by stretching the
graph of 3y x vertically by a factor of 12
,
then shifting the resulting graph down four units.
82. 312
2g x x
This graph may be obtained by stretching the
graph of 3y x vertically by a factor of 12
,
then shifting the resulting graph up two units.
Section 2.7 Graphing Techniques 235
Copyright © 2017 Pearson Education, Inc.
83. 33g x x
This graph may be obtained by shifting the
graph of 3y x three units left.
84. 32f x x
This graph may be obtained by shifting the
graph of 3y x two units right.
85. 223
2f x x
This graph may be obtained by translating the
graph of 2y x two units to the right, then stretching the resulting graph vertically by a
factor of 23
.
86. Because ( ) ( ),g x x x f x the graphs
are the same.
87. (a) y = g(–x) The graph of g(x) is reflected across the y-axis.
(b) y = g(x – 2) The graph of g(x) is translated to the right 2 units.
(c) y = –g(x) The graph of g(x) is reflected across the x-axis.
(d) y = –g(x) + 2 The graph of g(x) is reflected across the x-axis and translated 2 units up.
236 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
88. (a) y f x
The graph of f(x) is reflected across the x-axis.
(b) 2y f x
The graph of f(x) is stretched vertically by a factor of 2.
(c) y f x
The graph of f(x) is reflected across the y-axis.
(d) 12
y f x
The graph of f(x) is compressed vertically
by a factor of 12
.
89. It is the graph of f x x translated 1 unit to
the left, reflected across the x-axis, and translated 3 units up. The equation is
1 3.y x
90. It is the graph of g x x translated 4 units
to the left, reflected across the x-axis, and translated two units up. The equation is
4 2.y x
91. It is the graph of f x x translated one
unit right and then three units down. The
equation is 1 3.y x
92. It is the graph of f x x translated 2 units
to the right, shrunken vertically by a factor of 12
, and translated one unit down. The
equation is 12
2 1.y x
93. It is the graph of g x x translated 4 units
to the left, stretched vertically by a factor of 2, and translated four units down. The equation
is 2 4 4.y x
94. It is the graph of f x x reflected across
the x-axis and then shifted two units down. The equation is 2y x .
95. Because f(3) = 6, the point (3, 6) is on the graph. Because the graph is symmetric with respect to the origin, the point (–3, –6) is on the graph. Therefore, f(–3) = –6.
96. Because f(3) = 6, (3, 6) is a point on the graph. The graph is symmetric with respect to the y-axis, so (–3, 6) is on the graph. Therefore, f(–3) = 6.
97. Because f(3) = 6, the point (3, 6) is on the graph. The graph is symmetric with respect to the line x = 6 and the point (3, 6) is 3 units to the left of the line x = 6, so the image point of (3, 6), 3 units to the right of the line x = 6 is (9, 6). Therefore, f(9) = 6.
98. Because f(3) = 6 and f(–x) = f(x), f(–3) = f(3). Therefore, f(–3) = 6.
99. Because (3, 6) is on the graph, (–3, –6) must also be on the graph. Therefore, f(–3) = –6.
100. If f is an odd function, f(–x) = –f(x). Because f(3) = 6 and f(–x) = –f(x), f(–3) = –f(3). Therefore, f(–3) = –6.
Chapter 2 Quiz (Sections 2.5−2.7) 237
Copyright © 2017 Pearson Education, Inc.
101. f(x) = 2x + 5 Translate the graph of ( )f x up 2 units to obtain the graph of
( ) (2 5) 2 2 7.t x x x Now translate the graph of t(x) = 2x + 7 left 3 units to obtain the graph of
( ) 2( 3) 7 2 6 7 2 13.g x x x x (Note that if the original graph is first translated to the left 3 units and then up 2 units, the final result will be the same.)
102. f(x) = 3 – x Translate the graph of ( )f x down 2 units to
obtain the graph of ( ) (3 ) 2 1.t x x x
Now translate the graph of 1t x x right
3 units to obtain the graph of ( ) ( 3) 1 3 1 4.g x x x x
(Note that if the original graph is first translated to the right 3 units and then down 2 units, the final result will be the same.)
103. (a) Because f(–x) = f(x), the graph is symmetric with respect to the y-axis.
(b) Because f(–x) = –f(x), the graph is symmetric with respect to the origin.
104. (a) f(x) is odd. An odd function has a graph symmetric with respect to the origin. Reflect the left half of the graph in the origin.
(b) f(x) is even. An even function has a graph symmetric with respect to the y-axis. Reflect the left half of the graph in the y-axis.
Chapter 2 Quiz (Sections 2.5−2.7)
1. (a) First, find the slope: 9 5
21 ( 3)
m
Choose either point, say, (−3, 5), to find the equation of the line:
5 2( ( 3)) 2( 3) 5y x y x 2 11y x .
(b) To find the x-intercept, let y = 0 and solve
for x: 112
0 2 11x x . The
x-intercept is 112
, 0 .
2. Write 3x − 2y = 6 in slope-intercept form to
find its slope: 32
3 2 6 3.x y y x
Then, the slope of the line perpendicular to
this graph is 23
. 23
4 ( ( 6))y x
2 23 3
( 6)) 4y x y x
3. (a) 8x (b) 5y
4. (a) Cubing function; domain: ( , ) ;
range: ( , ) ; increasing over
( , ) .
(b) Absolute value function; domain: ( , ) ; range: [0, ) ; decreasing over
( , 0) ; increasing over (0, )
238 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(c) Cube root function: domain: ( , ) ;
range: ( , ) ; increasing over
( , ) .
5.
0.40 0.75
5.5 0.40 5.5 0.75
0.40 5 0.75 2.75
f x xf
A 5.5-minute call costs $2.75.
6. if 0( )2 3 if 0
x xf xx x
For values of x < 0, the graph is the line y = 2x + 3. Do not include the right endpoint
(0, 3). Graph the line y x for values of x ≥ 0, including the left endpoint (0, 0).
7. 3( ) 1f x x
Reflect the graph of 3( )f x x across the x-axis, and then translate the resulting graph one unit up.
8. ( ) 2 1 3f x x
Shift the graph of ( )f x x one unit right,
stretch the resulting graph vertically by a factor of 2, then shift this graph three units up.
9. This is the graph of ( )g x x , translated four units to the left, reflected across the x-axis, and then translated two units down.
The equation is 4 2y x .
10. (a) 2 7f x x
Replace x with –x to obtain
2
2
( ) 7
7
f x xf x x f x
The result is the same as the original function, so the function is even.
(b) 3 1f x x x
Replace x with –x to obtain
3
3
1
1
f x x xx x f x
The result is not the same as the original equation, so the function is not even. Because ,f x f x the function is
not odd. Therefore, the function is neither even nor odd.
(c) 101 99f x x x
Replace x with –x to obtain
101 99
101 99
101 99
f x x xx x
x xf x
Because f(−x) = −f(x), the function is odd.
Section 2.8 Function Operations and Composition
In exercises 1–10, 1f x x and 2.g x x
1. 2
2 2 2
2 1 2 7
f g f g
2. 2
2 2 2
2 1 2 1
f g f g
3. 2
2 2 2
2 1 2 12
fg f g
4. 2
2 2 1 32
2 42
ffg g
5. 2 22 2 2 2 1 5f g f g f
Section 2.8 Function Operations and Composition 239
Copyright © 2017 Pearson Education, Inc.
6. 22 2 2 1 2 1 9g f g f g
7. f is defined for all real numbers, so its domain is , .
8. g is defined for all real numbers, so its domain is , .
9. f + g is defined for all real numbers, so its domain is , .
10. fg
is defined for all real numbers except those
values that make 0,g x so its domain is
, 0 0, .
In Exercises 11–18, 2( ) 3f x x and
( ) 2 6g x x .
11.
2
( )(3) (3) (3)
(3) 3 2(3) 6
12 0 12
f g f g
12. 2
( )( 5) ( 5) ( 5)
[( 5) 3] [ 2( 5) 6]28 16 44
f g f g
13. 2
( )( 1) ( 1) ( 1)
[( 1) 3] [ 2( 1) 6]4 8 4
f g f g
14. 2
( )(4) (4) (4)
[(4) 3] [ 2(4) 6]19 ( 2) 21
f g f g
15. 2
( )(4) (4) (4)
[4 3] [ 2(4) 6]19 ( 2) 38
fg f g
16. 2
( )( 3) ( 3) ( 3)
[( 3) 3] [ 2( 3) 6]12 12 144
fg f g
17. 2( 1) ( 1) 3 4 1
( 1)( 1) 2( 1) 6 8 2
f fg g
18. 2(5) (5) 3 28
(5) 7(5) 2(5) 6 4
f fg g
19. ( ) 3 4, ( ) 2 5f x x g x x
( )( ) ( ) ( )(3 4) (2 5) 5 1
f g x f x g xx x x
( )( ) ( ) ( )(3 4) (2 5) 9
f g x f x g xx x x
2
2
( )( ) ( ) ( ) (3 4)(2 5)
6 15 8 20
6 7 20
fg x f x g x x xx x xx x
( ) 3 4( )
( ) 2 5
f f x xxg g x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g,
and fg are all , . The domain of fg is
the set of all real numbers for which 0.g x This is the set of all real numbers
except 52
, which is written in interval notation
as 5 52 2
, , .
20. f(x) = 6 – 3x, g(x) = –4x + 1 ( )( ) ( ) ( )
(6 3 ) ( 4 1)7 7
f g x f x g xx x
x
( )( ) ( ) ( )(6 3 ) ( 4 1) 5
f g x f x g xx x x
2
2
( )( ) ( ) ( ) (6 3 )( 4 1)
24 6 12 3
12 27 6
fg x f x g x x xx x x
x x
( ) 6 3( )
( ) 4 1
f f x xxg g x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all , . The domain of
fg is the set of all real numbers for which
0.g x This is the set of all real numbers
except 14
, which is written in interval notation
as 1 14 4
– , , .
21. 2 2( ) 2 3 , ( ) 3f x x x g x x x
2 2
2
( )( ) ( ) ( )
(2 3 ) ( 3)
3 4 3
f g x f x g xx x x x
x x
2 2
2 2
2
( )( ) ( ) ( )
(2 3 ) ( 3)
2 3 3
2 3
f g x f x g xx x x x
x x x xx x
(continued on next page)
240 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
2 2
4 3 2 3 2
4 3 2
( )( ) ( ) ( )
(2 3 )( 3)
2 2 6 3 3 9
2 5 9 9
fg x f x g xx x x x
x x x x x xx x x x
2
2
( ) 2 3( )
( ) 3
f f x x xxg g x x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all , . The domain of
fg is the set of all real numbers for which
0.g x If 2 3 0x x , then by the
quadratic formula 1 112ix . The equation
has no real solutions. There are no real numbers which make the denominator zero.
Thus, the domain of fg is also , .
22. 2 2( ) 4 2 , ( ) 3 2f x x x g x x x
2 2
2
( )( ) ( ) ( )
(4 2 ) ( 3 2)
5 2
f g x f x g xx x x x
x x
2 2
2 2
2
( )( ) ( ) ( )
(4 2 ) ( 3 2)
4 2 3 2
3 5 2
f g x f x g xx x x x
x x x xx x
2 2
4 3 2 3 2
4 3 2
( )( ) ( ) ( )
(4 2 )( 3 2)
4 12 8 2 6 4
4 10 2 4
fg x f x g xx x x x
x x x x x xx x x x
2
2
( ) 4 2( )
( ) 3 2
f f x x xxg g x x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g,
and fg are all , . The domain of fg is the
set of all real numbers x such that 2 3 2 0x x . Because 2 3 2 ( 1)( 2)x x x x , the numbers
which give this denominator a value of 0 are
x = 1 and x = 2. Therefore, the domain of fg is
the set of all real numbers except 1 and 2, which is written in interval notation as (– , 1) (1, 2) (2, ) .
23. 1
( ) 4 1, ( )f x x g xx
1( )( ) ( ) ( ) 4 1f g x f x g x x
x
1( )( ) ( ) ( ) 4 1f g x f x g x x
x
( )( ) ( ) ( )
1 4 14 1
fg x f x g xxx
x x
1
( ) 4 1( ) 4 1
( ) x
f f x xx x xg g x
Because 14
4 1 0 4 1 ,x x x the
domain of f is 14
, . The domain of g is
, 0 0, . Considering the intersection
of the domains of f and g, the domains of f + g,
f – g, and fg are all 14
, . Because 1 0x
for any value of x, the domain of fg is also
14
, .
24. 1
( ) 5 4, ( )f x x g xx
( )( ) ( ) ( )1 1
5 4 5 4
f g x f x g x
x xx x
( )( ) ( ) ( )1 1
5 4 5 4
f g x f x g x
x xx x
( )( ) ( ) ( )
1 5 45 4
fg x f x g xxx
x x
1
( ) 5 4( ) 5 4
( ) x
f f x xx x xg g x
Because 45
5 4 0 5 4 ,x x x the
domain of f is 45
, . The domain of g is
, 0 0, . Considering the intersection
of the domains of f and g, the domains of f + g,
f – g, and fg are all 45
, . 1 0x for any
value of x, so the domain of fg is also
45
, .
Section 2.8 Function Operations and Composition 241
Copyright © 2017 Pearson Education, Inc.
25. 2008 280 and 2008 470, thus
2008 2008 2008280 470 750 (thousand).
M FT M F
26. 2012 390 and 2012 630, thus
2012 2012 2012390 630 1020 (thousand).
M FT M F
27. Looking at the graphs of the functions, the slopes of the line segments for the period 2008−2012 are much steeper than the slopes of the corresponding line segments for the period 2004−2008. Thus, the number of associate’s degrees increased more rapidly during the period 2008−2012.
28. If 2004 2012, ( ) ,k T k r and F(k) = s, then M(k) = r − s.
29. 2000 2000 200019 13 6
T S T S
It represents the dollars in billions spent for general science in 2000.
30. 2010 2010 201029 11 18
T G T G
It represents the dollars in billions spent on space and other technologies in 2010.
31. Spending for space and other technologies spending decreased in the years 1995−2000 and 2010–2015.
32. Total spending increased the most during the years 2005−2010.
33. (a)
2 2 2
4 2 2
f g f g
(b) ( )(1) (1) (1) 1 ( 3) 4f g f g
(c) ( )(0) (0) (0) 0( 4) 0fg f g
(d) (1) 1 1
(1)(1) 3 3
f fg g
34. (a) ( )(0) (0) (0) 0 2 2f g f g
(b) ( )( 1) ( 1) ( 1)2 1 3
f g f g
(c) ( )(1) (1) (1) 2 1 2fg f g
(d) (2) 4
(2) 2(2) 2
f fg g
35. (a) ( )( 1) ( 1) ( 1) 0 3 3f g f g
(b) ( )( 2) ( 2) ( 2)1 4 5
f g f g
(c) ( )(0) (0) (0) 1 2 2fg f g
(d) (2) 3
(2) undefined(2) 0
f fg g
36. (a) ( )(1) (1) (1) 3 1 2f g f g
(b) ( )(0) (0) (0) 2 0 2f g f g
(c) ( )( 1) ( 1) ( 1) 3( 1) 3fg f g
(d) (1) 3
(1) 3(1) 1
f fg g
37. (a) 2 2 2 7 2 5f g f g
(b) ( )(4) (4) (4) 10 5 5f g f g
(c) ( )( 2) ( 2) ( 2) 0 6 0fg f g
(d) (0) 5
(0) undefined(0) 0
f fg g
38. (a) 2 2 2 5 4 9f g f g
(b) ( )(4) (4) (4) 0 0 0f g f g
(c) ( )( 2) ( 2) ( 2) 4 2 8fg f g
(d) (0) 8
(0) 8(0) 1
f fg g
242 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
39. x f x g x f g x f g x fg x f xg
−2 0 6 0 6 6 0 6 6 0 6 0 06
0
0 5 0 5 0 5 5 0 5 5 0 0 50
undefined
2 7 −2 7 2 5 7 2 9 7 2 14 72
3.5
4 10 5 10 5 15 10 5 5 10 5 50 105
2
40. x f x g x f g x f g x fg x f xg
−2 −4 2 4 2 2 4 2 6 4 2 8 42
2
0 8 −1 8 1 7 8 1 9 8 1 8 81
8
2 5 4 5 4 9 5 4 1 5 4 20 54
1.25
4 0 0 0 0 0 0 0 0 0 0 0 00
undefined
41. Answers may vary. Sample answer: Both the
slope formula and the difference quotient represent the ratio of the vertical change to the horizontal change. The slope formula is stated for a line while the difference quotient is stated for a function f.
42. Answers may vary. Sample answer: As h approaches 0, the slope of the secant line PQ approaches the slope of the line tangent of the curve at P.
43. 2f x x
(a) ( ) 2 ( ) 2f x h x h x h
(b) ( ) ( ) (2 ) (2 )2 2
f x h f x x h xx h x h
(c) ( ) ( )
1f x h f x h
h h
44. 1f x x
(a) ( ) 1 ( ) 1f x h x h x h
(b) ( ) ( ) (1 ) (1 )1 1
f x h f x x h xx h x h
(c) ( ) ( )
1f x h f x h
h h
45. 6 2f x x
(a) ( ) 6( ) 2 6 6 2f x h x h x h
(b) ( ) ( )(6 6 2) (6 2)6 6 2 6 2 6
f x h f xx h x
x h x h
(c) ( ) ( ) 6
6f x h f x h
h h
46. 4 11f x x
(a) ( ) 4( ) 11 4 4 11f x h x h x h
(b) ( ) ( )(4 4 11) (4 11)4 4 11 4 11 4
f x h f xx h x
x h x h
(c) ( ) ( ) 4
4f x h f x h
h h
47. 2 5f x x
(a) ( ) 2( ) 52 2 5
f x h x hx h
(b) ( ) ( )( 2 2 5) ( 2 5)
2 2 5 2 5 2
f x h f xx h x
x h x h
(c) ( ) ( ) 2
2f x h f x h
h h
Section 2.8 Function Operations and Composition 243
Copyright © 2017 Pearson Education, Inc.
48. ( ) 4 2f x x
(a) ( ) 4( ) 24 4 2
f x h x hx h
(b)
( ) ( )4 4 2 4 24 4 2 4 24
f x h f xx h xx h xh
(c) ( ) ( ) 4
4f x h f x h
h h
49. 1
( )f xx
(a) 1
( )f x hx h
(b)
( ) ( )
1 1
f x h f xx x h
x h x x x hh
x x h
(c)
( ) ( )
1
hx x hf x h f x h
h h hx x h
x x h
50. 2
1( )f x
x
(a) 2
1( )f x h
x h
(b)
22
2 2 22
2 2 2 2
2 22 2
( ) ( )
1 1
2 2
f x h f xx x h
xx h x x hx x xh h xh h
x x h x x h
(c)
2
22
22
22
22
22
( ) ( ) 2
2
2
xh hx x hf x h f x xh h
h h hx x hh x h
hx x hx h
x x h
51. 2( )f x x
(a) 2 2 2( ) ( ) 2f x h x h x xh h
(b) 2 2 2
2( ) ( ) 2
2
f x h f x x xh h xxh h
(c) 2( ) ( ) 2
(2 )
2
f x h f x xh hh h
h x hh
x h
52. 2( )f x x
(a)
2
2 2
2 2
( ) ( )
2
2
f x h x hx xh h
x xh h
(b) 2 2 2
2 2 2
2
( ) ( ) 2
2
2
f x h f x x xh h x
x xh h xxh h
(c) 2( ) ( ) 2
(2 )
2
f x h f x xh hh h
h x hh
x h
53. 2( ) 1f x x
(a) 2
2 2
2 2
( ) 1 ( )
1 ( 2 )
1 2
f x h x hx xh h
x xh h
(b) 2 2 2
2 2 2
2
( ) ( )
(1 2 ) (1 )
1 2 1
2
f x h f xx xh h x
x xh h xxh h
(c) 2( ) ( ) 2
( 2 )
2
f x h f x xh hh h
h x hh
x h
54. 2( ) 1 2f x x
(a) 2
2 2
2 2
( ) 1 2( )
1 2( 2 )
1 2 4 2
f x h x hx xh h
x xh h
244 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b)
2 2 2
2 2 2
2
( ) ( )
1 2 4 2 1 2
1 2 4 2 1 2
4 2
f x h f xx xh h x
x xh h xxh h
(c) 2( ) ( ) 4 2
(4 2 )
4 2
f x h f x xh hh h
h x hh
x h
55. 2( ) 3 1f x x x
(a) 2
2 2
( ) 3 1
2 3 3 1
f x h x h x hx xh h x h
(b)
2 2
2
2 2 2
2
( ) ( )
2 3 3 1
3 1
2 3 3 1 3 1
2 3
f x h f xx xh h x h
x x
x xh h x h x xxh h h
(c) 2( ) ( ) 2 3
(2 3)
2 3
f x h f x xh h hh h
h x hh
x h
56. 2( ) 4 2f x x x
(a) 2
2 2
( ) 4 2
2 4 4 2
f x h x h x hx xh h x h
(b)
2 2
2
2 2 2
2
( ) ( )
2 4 4 2
4 2
2 4 4 2 4 2
2 4
f x h f xx xh h x h
x x
x xh h x h x xxh h h
(c) 2( ) ( ) 2 4
(2 4)
2 4
f x h f x xh h hh h
h x hh
x h
57. 3 4 4 3 1g x x g
4 4 1
2 1 3 2 3 5
f g f g f
58. 3 2 2 3 1g x x g
2 2 1
2 1 3 2 3 1
f g f g f
59. 3 2 2 3 5g x x g
2 2 5
2 5 3 10 3 7
f g f g f
60. 2 3f x x 3 2 3 3 6 3 3f
3 3 3 3 3 0g f g f g
61. 2 3 0 2 0 3 0 3 3f x x f
0 0 3
3 3 3 3 6
g f g f g
62. 2 3 2 2 2 3 7f x x f
2 2 7
7 3 7 3 10
g f g f g
63. 2 3 2 2 2 3 4 3 1f x x f
2 2 1 2 1 3 1f f f f f
64. 3 2 2 3 5g x x g
2 2 5 5 3 2g g g g g
65. ( )(2) [ (2)] (3) 1f g f g f
66. ( )(7) [ (7)] (6) 9f g f g f
67. ( )(3) [ (3)] (1) 9g f g f g
68. ( )(6) [ (6)] (9) 12g f g f g
69. ( )(4) [ (4)] (3) 1f f f f f
70. ( )(1) [ (1)] (9) 12g g g g g
71. ( )(1) [ (1)] (9)f g f g f However, f(9) cannot be determined from the table given.
72. ( ( ))(7) ( ( (7)))( (6)) (9) 12
g f g g f gg f g
73. (a) ( )( ) ( ( )) (5 7)6(5 7) 930 42 9 30 33
f g x f g x f xx
x x
The domain and range of both f and g are ( , ) , so the domain of f g is
( , ) .
Section 2.8 Function Operations and Composition 245
Copyright © 2017 Pearson Education, Inc.
(b) ( )( ) ( ( )) ( 6 9)5( 6 9) 7
30 45 7 30 52
g f x g f x g xx
x x
The domain of g f is ( , ) .
74. (a) ( )( ) ( ( )) (3 1)8(3 1) 1224 8 12 24 4
f g x f g x f xxx x
The domain and range of both f and g are ( , ) , so the domain of f g is
( , ) .
(b) ( )( ) ( ( )) (8 12)3(8 12) 124 36 1 24 35
g f x g f x g xxx x
The domain of g f is ( , ) .
75. (a) ( )( ) ( ( )) ( 3) 3f g x f g x f x x
The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 3 0 3x x .
Therefore, the domain of f g is
[ 3, ) .
(b) ( )( ) ( ( )) 3g f x g f x g x x
The domain and range of g are ( , ) , however, the domain and range of f are [0, ) .Therefore, the domain of g f is
[0, ) .
76. (a) ( )( ) ( ( )) ( 1) 1f g x f g x f x x
The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 1 0 1x x . Therefore,
the domain of f g is [1, ) .
(b) ( )( ) ( ( )) 1g f x g f x g x x
The domain and range of g are ( , ), however, the domain and range of f are [0, ). Therefore, the domain of g f is
[0, ).
77. (a) 2
2 3
( )( ) ( ( )) ( 3 1)
( 3 1)
f g x f g x f x xx x
The domain and range of f and g are ( , ), so the domain of f g is
( , ).
(b)
3
23 3
6 3
( )( ) ( ( )) ( )
3 1
3 1
g f x g f x g x
x x
x x
The domain and range of f and g are ( , ), so the domain of g f is
( , ).
78. (a) 4 2
4 2
4 2
( )( ) ( ( )) ( 4)
4 2
2
f g x f g x f x xx xx x
The domain of f and g is ( , ), while
the range of f is ( , ) and the range of
g is 4, , so the domain of f g is
( , ).
(b) 4 2
( )( ) ( ( )) ( 2)
( 2) ( 2) 4
g f x g f x g xx x
The domain of f and g is ( , ), while
the range of f is ( , ) and the range of
g is 4, , so the domain of g f is
( , ).
79. (a) ( )( ) ( ( )) (3 ) 3 1f g x f g x f x x
The domain and range of g are ( , ) ,
however, the domain of f is [1, ), while
the range of f is [0, ). So, 13
3 1 0x x . Therefore, the
domain of f g is 13
, .
(b) ( )( ) ( ( )) 1
3 1
g f x g f x g x
x
The domain and range of g are ( , ) ,
however, the range of f is [0, ) . So
1 0 1x x . Therefore, the domain of g f is [1, ) .
80. (a) ( )( ) ( ( )) (2 ) 2 2f g x f g x f x x
The domain and range of g are ( , ) ,
however, the domain of f is [2, ) . So,
2 2 0 1x x . Therefore, the
domain of f g is 1, .
246 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b) ( )( ) ( ( )) 2g f x g f x g x
2 2x
The domain and range of g are ( , ) ,
however, the range of f is [0, ) . So
2 0 2x x . Therefore, the domain of g f is [2, ) .
81. (a) 21
( )( ) ( ( )) ( 1) xf g x f g x f x
The domain and range of g are ( , ) , however, the domain of f is ( ,0) (0, ) . So, 1 0 1x x .
Therefore, the domain of f g is
( , 1) ( 1, ) .
(b) 2 2( )( ) ( ( )) 1x xg f x g f x g
The domain and range of f is ( ,0) (0, ) , however, the domain
and range of g are ( , ) . So 0x .
Therefore, the domain of g f is
( ,0) (0, ) .
82. (a) 44
( )( ) ( ( )) ( 4) xf g x f g x f x
The domain and range of g are ( , ) , however, the domain of f is ( ,0) (0, ) . So, 4 0 4x x .
Therefore, the domain of f g is
( , 4) ( 4, ) .
(b) 4 4( )( ) ( ( )) 4x xg f x g f x g
The domain and range of f is ( ,0) (0, ) , however, the domain
and range of g are ( , ) . So 0x .
Therefore, the domain of g f is
( ,0) (0, ) .
83. (a) 1 1( )( ) ( ( )) 2x xf g x f g x f
The domain and range of g are ( , 0) (0, ) , however, the domain
of f is [ 2, ) . So, 1 2 0x
12
0 or x x (using test intervals).
Therefore, the domain of f g is
12
, 0 , .
(b) 12
( )( ) ( ( )) 2x
g f x g f x g x
The domain of f is [ 2, ) and its range is
[0, ). The domain and range of g
are ( ,0) (0, ) . So 2 0 2x x .
Therefore, the domain of g f is ( 2, ) .
84. (a) 2 2( )( ) ( ( )) 4x xf g x f g x f
The domain and range of g are ( , 0) (0, ) , however, the domain
of f is [ 4, ) . So, 2 4 0x
12
0 or x x (using test intervals).
Therefore, the domain of f g is
12
, 0 , .
(b) 24
( )( ) ( ( )) 4x
g f x g f x g x
The domain of f is [ 4, ) and its range is
[0, ). The domain and range of g
are ( , 0) (0, ) . So 4 0 4x x .
Therefore, the domain of g f is ( 4, ) .
85. (a) 1 15 5
( )( ) ( ( )) x xf g x f g x f
The domain of g is ( , 5) ( 5, ) ,
and the range of g is ( , 0) (0, ) .
The domain of f is [0, ) . Therefore, the
domain of f g is 5, .
(b) 15
( )( ) ( ( ))x
g f x g f x g x
The domain and range of f is [0, ). The
domain of g is ( , 5) ( 5, ).
Therefore, the domain of g f is [0, ).
86. (a) 3 36 6
( )( ) ( ( )) x xf g x f g x f
The domain of g is ( , 6) ( 6, ) ,
and the range of g is ( , 0) (0, ) .
The domain of f is [0, ) . Therefore, the
domain of f g is 6, .
(b) 36
( )( ) ( ( ))x
g f x g f x g x
The domain and range of f is [0, ) . The
domain of g is ( , 6) ( 6, ) .
Therefore, the domain of g f is [0, ).
Section 2.8 Function Operations and Composition 247
Copyright © 2017 Pearson Education, Inc.
87. (a) 1 11 2 1 2
( )( ) ( ( )) xx x xf g x f g x f
The domain and range of g are ( , 0) (0, ) . The domain of f is
( , 2) ( 2, ), and the range of f is
( , 0) (0, ) . So, 1 2
0 0xx x or
12
0 x or 12
x (using test intervals).
Thus, 0x and 12
x . Therefore, the
domain of f g is
1 12 2
, 0 0, , .
(b) 1 12 1 ( 2)
( )( ) ( ( ))
2x xg f x g f x g
x
The domain and range of g are ( , 0) (0, ). The domain of f is
( , 2) (2, ), and the range of f is
( , 0) (0, ). Therefore, the domain of
g f is ( , 2) (2, ).
88. (a) 1 11 4
1 4
( )( ) ( ( )) x xx
x
f g x f g x f
The domain and range of g are ( , 0) (0, ). The domain of f is
( , 4) ( 4, ), and the range of f is
( , 0) (0, ). So, 1 4
0 0xx x
or 14
0 x or 14
1 4 0x x
(using test intervals). Thus, x ≠ 0 and
x ≠ 14
. Therefore, the domain of f g is
1 14 4
, 0 0, , .
(b) 1 14 1 ( 4)
( )( ) ( ( ))
4x xg f x g f x g
x
The domain and range of g are ( ,0) (0, ) . The domain of f is
( , 4) ( 4, ) , and the range of f is
( ,0) (0, ) . Therefore, the domain of
g f is ( , 4) ( 4, ).
89.
(2) (1) 2 and (3) (2) 5
Since (1) 7 and (1) 3, (3) 7.
g f g g f gg f f g
x f x g x ( )g f x
1 3 2 7
2 1 5 2
3 2 7 5
90. ( )f x is odd, so ( 1) (1) ( 2) 2.f f
Because ( )g x is even, (1) ( 1) 2g g and
(2) ( 2) 0.g g ( )( 1) 1,f g so
( 1) 1f g and (2) 1.f ( )f x is odd, so
( 2) (2) 1.f f Thus,
( )( 2) ( 2) (0) 0f g f g f and
( )(1) (1) (2) 1f g f g f and
( )(2) (2) (0) 0.f g f g f
x –2 –1 0 1 2
f x –1 2 0 –2 1
g x 0 2 1 2 0
f g x 0 1 –2 1 0
91. Answers will vary. In general, composition of functions is not commutative. Sample answer:
2 3 3 2 3 26 9 2 6 11
3 2 2 3 2 36 4 3 6 7
f g x f x xx x
g f x g x xx x
Thus, f g x is not equivalent to
g f x .
92.
3
33
7
7 7
7 7
f g x f g x f x
xx x
3
33 33
7
7 7
g f x g f x g x
x x x
93.
14
14
4 2 2
4 2 2
2 2 2 2
f g x f g x x
xx x x
14
1 14 4
4 2 2
4 2 2 4
g f x g f x xx x x
94.
13
13
3
3
f g x f g x x
x x
13
13
3
3
g f x g f x x
x x
248 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
95. 31 435 5
3 33 3
5 4
4 4
f g x f g x x
x x x
31 435 5
51 4 4 45 5 5 5 555
5 4
5 4 x
x
g f x g f x x
xx
96. 33
3 33 3
1 1
1 1
f g x f g x x
x x x
33 1 1
1 1
g f x g f x xx x
In Exercises 97−102, we give only one of many possible answers.
97. 2( ) (6 2)h x x
Let g(x) = 6x – 2 and 2( )f x x . 2( )( ) (6 2) (6 2) ( )f g x f x x h x
98. 2 2( ) (11 12 )h x x x 2 2Let ( ) 11 12 and ( ) .g x x x f x x
2
2 2
( )( ) (11 12 )
(11 12 ) ( )
f g x f x xx x h x
99. 2( ) 1h x x
Let 2( ) 1 and ( ) .g x x f x x
2 2( )( ) ( 1) 1 ( ).f g x f x x h x
100. 3( ) (2 3)h x x 3Let ( ) 2 3 and ( ) .g x x f x x
3( )( ) (2 3) (2 3) ( )f g x f x x h x
101. ( ) 6 12h x x
Let ( ) 6 and ( ) 12.g x x f x x
( )( ) (6 ) 6 12 ( )f g x f x x h x
102. 3( ) 2 3 4h x x
Let 3( ) 2 3 and ( ) 4.g x x f x x 3( )( ) (2 3) 2 3 4 ( )f g x f x x h x
103. f(x) = 12x, g(x) = 5280x ( )( ) [ ( )] (5280 )
12(5280 ) 63,360f g x f g x f x
x x
The function f g computes the number of inches in x miles.
104. 3 , 1760f x x g x x
1760
3 1760 5280
f g x f g x f xx x
f g x compute the number of feet in x
miles.
105. 23( )
4x x
(a) 2 2 23 3(2 ) (2 ) (4 ) 3
4 4x x x x
(b) 2(16) (2 8) 3(8)
64 3 square units
A
106. (a) 4 4
4 x xx s s s
(b) 2224 16x xy s
(c) 26 36
2.25 square units16 16
y
107. (a) 2( ) 4 and ( )r t t r r
2 2( )( ) [ ( )]
(4 ) (4 ) 16
r t r tt t t
(b) ( )( )r t defines the area of the leak in terms of the time t, in minutes.
(c) 2 2(3) 16 (3) 144 ft
108. (a) 2 2
( )( ) [ ( )]
(2 ) (2 ) 4
r t r tt t t
(b) It defines the area of the circular layer in terms of the time t, in hours.
(c) 2 2( )(4) 4 (4) 64 mir
109. Let x = the number of people less than 100 people that attend.
(a) x people fewer than 100 attend, so 100 – x people do attend N(x) = 100 – x
(b) The cost per person starts at $20 and increases by $5 for each of the x people that do not attend. The total increase is $5x, and the cost per person increases to $20 + $5x. Thus, G(x) = 20 + 5x.
(c) ( ) ( ) ( ) (100 )(20 5 )C x N x G x x x
Chapter 2 Review Exercises 249
Copyright © 2017 Pearson Education, Inc.
(d) If 80 people attend, x = 100 – 80 = 20.
20 100 20 20 5 20
80 20 100
80 120 $9600
C
110. (a) 1 0.02y x
(b) 2 0.015( 500)y x
(c) 1 2y y represents the total annual interest.
(d) 1 2
1 2
( )(250)(250) (250)
0.02(250) 0.015(250 500)5 0.015(750) 15 11.25$16.25
y yy y
111. (a) 1
2g x x
(b) 1f x x
(c) 1 11
2 2f g x f g x f x x
(d) 160 60 1 $31
2f g
112. If the area of a square is 2x square inches, each side must have a length of x inches. If 3 inches is added to one dimension and 1 inch is subtracted from the other, the new dimensions will be x + 3 and x – 1. Thus, the area of the resulting rectangle is (x) = (x + 3)(x – 1).
Chapter 2 Review Exercises
1. P(3, –1), Q(–4, 5) 2 2
2 2
( , ) ( 4 3) [5 ( 1)]
( 7) 6 49 36 85
d P Q
Midpoint:
3 ( 4) 1 5 1 4 1, , , 2
2 2 2 2 2
2. M(–8, 2), N(3, –7) 2 2
2 2
( , ) [3 ( 8)] ( 7 2)
11 ( 9) 121 81 202
d M N
Midpoint: 8 3 2 ( 7) 5 5
, , 2 2 2 2
3. A(–6, 3), B(–6, 8) 2 2
2
( , ) [ 6 ( 6)] (8 3)
0 5 25 5
d A B
Midpoint: 6 ( 6) 3 8 12 11 11
, , 6,2 2 2 2 2
4. Label the points A(5, 7), B(3, 9), and C(6, 8).
2 2
2 2
( , ) 3 5 9 7
2 2 4 4 8
d A B
2 2
2 2
( , ) (6 5) (8 7)
1 1 1 1 2
d A C
2 2
22
( , ) 6 3 8 9
3 1 9 1 10
d B C
Because 2 2 28 2 10 , triangle
ABC is a right triangle with right angle at (5, 7).
5. Let B have coordinates (x, y). Using the midpoint formula, we have
6 108, 2 ,
2 26 10
8 22 2
6 16 10 422 6
x y
x y
x yx y
The coordinates of B are (22, −6).
6. P(–2, –5), Q(1, 7), R(3, 15) 2 2
2 2
( , ) (1 ( 2)) (7 ( 5))
(3) (12) 9 144
153 3 17
d P Q
2 2
2 2
( , ) (3 1) (15 7)
2 8 4 64 68 2 17
d Q R
2 2
2 2
( , ) (3 ( 2)) (15 ( 5))
(5) (20) 25 400 5 17
d P R
(continued on next page)
250 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
( , ) ( , ) 3 17 2 17d P Q d Q R
5 17 ( , ),d P R so these three points are collinear.
7. Center (–2, 3), radius 15 2 2 2
2 2 2
2 2
( ) ( )
[ ( 2)] ( 3) 15
( 2) ( 3) 225
x h y k rx y
x y
8. Center ( 5, 7 ), radius 3
2 2 2
22 2
2 2
( ) ( )
5 7 3
5 7 3
x h y k r
x y
x y
9. Center (–8, 1), passing through (0, 16) The radius is the distance from the center to any point on the circle. The distance between (–8, 1) and (0, 16) is
2 2 2 2(0 ( 8)) (16 1) 8 15
64 225 289 17.
r
The equation of the circle is 2 2 2
2 2[ ( 8)] ( 1) 17
( 8) ( 1) 289
x yx y
10. Center (3, –6), tangent to the x-axis The point (3, –6) is 6 units directly below the x-axis. Any segment joining a circle’s center to a point on the circle must be a radius, so in this case the length of the radius is 6 units.
2 2 2
2 2 2
2 2
( ) ( )
( 3) [ ( 6)] 6
( 3) ( 6) 36
x h y k rx y
x y
11. The center of the circle is (0, 0). Use the distance formula to find the radius:
2 2 2(3 0) (5 0) 9 25 34r
The equation is 2 2 34x y .
12. The center of the circle is (0, 0). Use the distance formula to find the radius:
2 2 2( 2 0) (3 0) 4 9 13r
The equation is 2 2 13x y .
13. The center of the circle is (0, 3). Use the distance formula to find the radius:
2 2 2( 2 0) (6 3) 4 9 13r
The equation is 2 2( 3) 13x y .
14. The center of the circle is (5, 6). Use the distance formula to find the radius:
2 2 2(4 5) (9 6) 1 9 10r
The equation is 2 2( 5) ( 6) 10x y .
15. 2 24 6 12 0x x y y Complete the square on x and y to put the equation in center-radius form.
2 2
2 2
2 2
4 6 12
4 4 6 9 12 4 9
2 3 1
x x y y
x x y y
x y
The circle has center (2, –3) and radius 1.
16. 2 26 10 30 0x x y y Complete the square on x and y to put the equation in center-radius form.
2 2
2 2
( 6 9) ( 10 25) 30 9 25
( 3) ( 5) 4
x x y yx y
The circle has center (3, 5) and radius 2.
17.
2 2
2 2
2 2
2 249 9 49 94 4 4 4
2 27 3 49 942 2 4 4 4
2 27 3 542 2 4
2 14 2 6 2 0
7 3 1 0
7 3 1
7 3 1
x x y yx x y yx x y y
x x y y
x y
x y
The circle has center 7 32 2
, and radius
54 9 6 3 6544 24 4
.
18.
2 2
2 2
2 2
2 2 25 25121 1214 4 4 4
22 5 146112 2 4
3 33 3 15 0
11 5 0
11 5 0
11 5 0
x x y yx x y y
x x y y
x x y y
x y
The circle has center 5112 2
, and radius
1462
.
19. This is not the graph of a function because a vertical line can intersect it in two points. domain: [−6, 6]; range: [−6, 6]
20. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: 0,
Chapter 2 Review Exercises 251
Copyright © 2017 Pearson Education, Inc.
21. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: ( , 1] [1, )
22. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: 0,
23. This is not the graph of a function because a vertical line can intersect it in two points. domain: 0, ; range: ,
24. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: ,
25. 26y x Each value of x corresponds to exactly one value of y, so this equation defines a function.
26. The equation 213
x y does not define y as a
function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 3) and (3, –3) satisfy the relation. Thus, the relation would not be a function.
27. The equation 2y x does not define y as a function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 1) and (3, −1) satisfy the relation.
28. The equation 4yx
defines y as a function
of x because for every x in the domain, which is (– , 0) (0, ) , there will be exactly one value of y.
29. In the function ( ) 4f x x , we may use
any real number for x. The domain is , .
30. 8
( )8
xf xx
x can be any real number except 8 because this will give a denominator of zero. Thus, the domain is (– , 8) (8, ).
31. 6 3f x x
In the function 6 3y x , we must have
6 3 0x . 6 3 0 6 3 2 2x x x x Thus, the domain is , 2 .
32. (a) As x is getting larger on the interval 2, , the value of y is increasing.
(b) As x is getting larger on the interval , 2 , the value of y is decreasing.
(c) f(x) is constant on (−2, 2).
In exercises 33–36, 22 3 6.f x x x
33. 23 2 3 3 3 62 9 3 3 618 9 6 15
f
34.
20.5 2 0.5 3 0.5 6
2 0.25 3 0.5 60.5 1.5 6 8
f
35. 20 2 0 3 0 6 6f
36. 22 3 6f k k k
37. 25
2 5 5 5 2 5 1x y y x y x
The graph is the line with slope 25
and
y-intercept (0, –)1. It may also be graphed using intercepts. To do this, locate the x-intercept: 0y
52
2 5 0 5 2 5x x x
38. 37
3 7 14 7 3 14 2x y y x y x
The graph is the line with slope of 37
and
y-intercept (0, 2). It may also be graphed using intercepts. To do this, locate the x-intercept by setting y = 0:
143
3 7 0 14 3 14x x x
252 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
39. 25
2 5 20 5 2 20 4x y y x y x
The graph is the line with slope of 25
and
y-intercept (0, 4). It may also be graphed using intercepts. To do this, locate the x-intercept: x-intercept: 0y
2 5 0 20 2 20 10x x x
40. 13
3y x y x
The graph is the line with slope 13
and
y-intercept (0, 0), which means that it passes through the origin. Use another point such as (6, 2) to complete the graph.
41. f(x) = x The graph is the line with slope 1 and y-intercept (0, 0), which means that it passes through the origin. Use another point such as (1, 1) to complete the graph.
42.
14
4 84 8
2
x yy xy x
The graph is the line with slope 14
and
y-intercept (0, –2). It may also be graphed using intercepts. To do this, locate the x-intercept:
0 4 0 8 8y x x
43. x = –5 The graph is the vertical line through (–5, 0).
44. f(x) = 3 The graph is the horizontal line through (0, 3).
45. 2 0 2y y The graph is the horizontal line through (0, −2).
46. The equation of the line that lies along the x-axis is y = 0.
47. Line through (0, 5), 23
m
Note that 2 23 3
.m
Begin by locating the point (0, 5). Because the
slope is 23 , a change of 3 units horizontally
(3 units to the right) produces a change of –2 units vertically (2 units down). This gives a second point, (3, 3), which can be used to complete the graph.
(continued on next page)
Chapter 2 Review Exercises 253
Copyright © 2017 Pearson Education, Inc.
(continued)
48. Line through (2, –4), 34
m
First locate the point (2, –4).
Because the slope is 34
, a change of 4 units
horizontally (4 units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (6, –1), which can be used to complete the graph.
49. through (2, –2) and (3, –4) 2 1
2 1
4 2 22
3 2 1
y ymx x
50. through (8, 7) and 12
, 2
2 11 15
2 1 2 2
2 7 9
8
2 18 69
15 15 5
y ymx x
51. through (0, –7) and (3, –7) 7 7 0
03 0 3
m
52. through (5, 6) and (5, –2)
2 1
2 1
2 6 8
5 5 0
y ymx x
The slope is undefined.
53. 11 2 3x y Solve for y to put the equation in slope-intercept form.
3112 2
2 11 3y x y x
Thus, the slope is 112
.
54. 9x – 4y = 2. Solve for y to put the equation in slope-intercept form.
9 14 2
4 9 2y x y x
Thus, the slope is 94
.
55. 2 0 2x x The graph is a vertical line, through (2, 0). The slope is undefined.
56. x – 5y = 0. Solve for y to put the equation in slope-intercept form.
15
5y x y x
Thus, the slope is 15
.
57. Initially, the car is at home. After traveling for 30 mph for 1 hr, the car is 30 mi away from home. During the second hour the car travels 20 mph until it is 50 mi away. During the third hour the car travels toward home at 30 mph until it is 20 mi away. During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home. During the last hour, the car travels 60 mi at 60 mph until it arrived home.
58. (a) This is the graph of a function because no vertical line intersects the graph in more than one point.
(b) The lowest point on the graph occurs in December, so the most jobs lost occurred in December. The highest point on the graph occurs in January, so the most jobs gained occurred in January.
(c) The number of jobs lost in December is approximately 6000. The number of jobs gained in January is approximately 2000.
(d) It shows a slight downward trend.
59. (a) We need to first find the slope of a line that passes between points (0, 30.7) and (12, 82.9)
2 1
2 1
82.9 30.7 52.24.35
12 0 12
y ymx x
Now use the point-intercept form with b = 30.7 and m = 4.35. y = 4.35x + 30.7 The slope, 4.35, indicates that the number of e-filing taxpayers increased by 4.35% each year from 2001 to 2013.
(b) For 2009, we evaluate the function for x = 8. y = 4.35(8) + 30.7 = 65.5 65.5% of the tax returns are predicted to have been filed electronically.
254 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
60. We need to find the slope of a line that passes between points (1980, 21000) and (2013, 63800)
2 1
2 1
63,800 21,000
2013 198042,800
$1297 per year33
y ymx x
The average rate of change was about $1297 per year.
61. (a) through (3, –5) with slope –2 Use the point-slope form.
1 1( )( 5) 2( 3)
5 2( 3)5 2 6
2 1
y y m x xy x
y xy x
y x
(b) Standard form: 2 1 2 1y x x y
62. (a) through (–2, 4) and (1, 3) First find the slope.
3 4 1
1 ( 2) 3m
Now use the point-slope form with 1
1 1 3( , ) (1, 3) and .x y m
1 113
1013 3
( )
3 ( 1)3( 3) 1( 1)
3 9 13 10
y y m x xy x
y xy x
y x y x
(b) Standard form: 101
3 33 10
3 10y x y xx y
63. (a) through (2, –1) parallel to 3x – y = 1 Find the slope of 3x – y = 1. 3 1 3 1 3 1x y y x y x The slope of this line is 3. Because parallel lines have the same slope, 3 is also the slope of the line whose equation is to be found. Now use the point-slope form with 1 1( , ) (2, 1) and 3.x y m
1 1( )( 1) 3( 2)
1 3 6 3 7
y y m x xy x
y x y x
(b) Standard form: 3 7 3 7 3 7y x x y x y
64. (a) x-intercept (–3, 0), y-intercept (0, 5) Two points of the line are (–3, 0) and (0, 5). First, find the slope.
5 0 5
0 3 3m
The slope is 53
and the y-intercept is
(0, 5). Write the equation in slope-
intercept form: 53
5y x
(b) Standard form: 53
5 3 5 155 3 15 5 3 15
y x y xx y x y
65. (a) through (2, –10), perpendicular to a line with an undefined slope A line with an undefined slope is a vertical line. Any line perpendicular to a vertical line is a horizontal line, with an equation of the form y = b. The line passes through (2, –10), so the equation of the line is y = –10.
(b) Standard form: y = –10
66. (a) through (0, 5), perpendicular to 8x + 5y = 3 Find the slope of 8x + 5y = 3.
8 35 5
8 5 3 5 8 3x y y xy x
The slope of this line is 85
. The slope
of any line perpendicular to this line is 58
, because 8 55 8
1.
The equation in slope-intercept form with
slope 58
and y-intercept (0, 5) is
58
5.y x
(b) Standard form: 58
5 8 5 405 8 40 5 8 40
y x y xx y x y
67. (a) through (–7, 4), perpendicular to y = 8 The line y = 8 is a horizontal line, so any line perpendicular to it will be a vertical line. Because x has the same value at all points on the line, the equation is x = –7. It is not possible to write this in slope-intercept form.
(b) Standard form: x = –7
68. (a) through (3, –5), parallel to y = 4 This will be a horizontal line through (3, –5). Because y has the same value at all points on the line, the equation is y = –5.
Chapter 2 Review Exercises 255
Copyright © 2017 Pearson Education, Inc.
(b) Standard form: y = –5
69. ( ) 3f x x
The graph is the same as that of ,y x
except that it is translated 3 units downward.
70. ( )f x x
The graph of ( )f x x is the reflection of
the graph of y x about the x-axis.
71. 2( ) 1 3f x x
The graph of 2( ) 1 3f x x is a
translation of the graph of 2y x to the left 1 unit, reflected over the x-axis and translated up 3 units.
72. ( ) 2f x x
The graph of ( ) 2f x x is the reflection
of the graph of y x about the x-axis, translated down 2 units.
73. ( ) 3f x x
To get y = 0, we need 0 3 1x 3 4.x To get y = 1, we need1 3 2x 4 5.x Follow this pattern to graph the step function.
74. 3( ) 2 1 2f x x
The graph of 3( ) 2 1 2f x x is a
translation of the graph of 3y x to the left 1 unit, stretched vertically by a factor of 2, and translated down 2 units.
75. 4 2 if 1( )
3 5 if 1x xf x x x
Draw the graph of y = –4x + 2 to the left of x = 1, including the endpoint at x = 1. Draw the graph of y = 3x – 5 to the right of x = 1, but do not include the endpoint at x = 1. Observe that the endpoints of the two pieces coincide.
256 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
76. 2 3 if 2( )
4 if 2x xf x
x x
Graph the curve 2 3y x to the left of x = 2, and graph the line y = –x + 4 to the right of x = 2. The graph has an open circle at (2, 7) and a closed circle at (2, 2).
77. if 3
( )6 if 3x xf x
x x
Draw the graph of y x to the left of x = 3,
but do not include the endpoint. Draw the graph of y = 6 – x to the right of x = 3, including the endpoint. Observe that the endpoints of the two pieces coincide.
78. Because x represents an integer, .x x
Therefore, 2 .x x x x x
79. True. The graph of an even function is symmetric with respect to the y-axis.
80. True. The graph of a nonzero function cannot be symmetric with respect to the x-axis. Such a graph would fail the vertical line test
81. False. For example, 2( )f x x is even and (2, 4) is on the graph but (2, –4) is not.
82. True. The graph of an odd function is symmetric with respect to the origin.
83. True. The constant function 0f x is both
even and odd. Because 0 ,f x f x
the function is even. Also 0 0 ,f x f x so the function is
odd.
84. False. For example, 3( )f x x is odd, and (2, 8) is on the graph but (–2, 8) is not.
85. 2 10x y
Replace x with –x to obtain 2( ) 10.x y The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to
obtain 2 2( ) 10 10.x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to
obtain 2 2( ) ( ) 10 ( ) 10.x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. The graph is symmetric with respect to the x-axis only.
86. 2 25 5 30y x Replace x with –x to obtain
2 2 2 25 5( ) 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2 2 25( ) 5 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the y-axis and x-axis, so it must also be symmetric with respect to the origin. Note that
this equation is the same as 2 2 6y x , which is a circle centered at the origin.
87. 2 3x y Replace x with –x to obtain
2 3 2 3( ) .x y x y The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace
y with –y to obtain 2 3 2 3( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y
with –y to obtain 2 3 2 3( ) ( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.
Chapter 2 Review Exercises 257
Copyright © 2017 Pearson Education, Inc.
88. 3 4y x
Replace x with –x to obtain 3 4y x . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain
3 3( ) 4 4y x y x 3 4y x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
3 3 3( ) ( ) 4 4 4.y x y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
89. 6 4x y
Replace x with –x to obtain 6( ) 4x y
6 4x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 6 ( ) 4 6 4x y x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain 6( ) ( ) 4x y 6 4x y . This equation is not equivalent to the original one, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
90. y x
Replace x with –x to obtain ( ) .y x y x The result is not the
same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain
.y x y x The result is the same as
the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
( ) .y x y x The result is not the
same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis only.
91. y = 1 This is the graph of a horizontal line through (0, 1). It is symmetric with respect to the y-axis, but not symmetric with respect to the x-axis and the origin.
92. x y
Replace x with –x to obtain .x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
.x y x y The result is the same as
the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.
93. 2 2 0x y Replace x with −x to obtain
2 2 2 20 0.x y x y The result is
the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain
22 2 20 0.x y x y The result is
the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.
94. 22 2 4x y
Replace x with −x to obtain
2 2 222 4 2 4.x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain
22 2 4.x y The result is not the same
as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
2 2 222 4 2 4,x y x y
which is not equivalent to the original equation. Therefore, the graph is not symmetric with respect to the origin.
95. To obtain the graph of ( )g x x , reflect the
graph of ( )f x x across the x-axis.
96. To obtain the graph of ( ) 2h x x , translate
the graph of ( )f x x down 2 units.
258 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
97. To obtain the graph of ( ) 2 4k x x ,
translate the graph of ( )f x x to the right 4
units and stretch vertically by a factor of 2.
98. If the graph of ( ) 3 4f x x is reflected about the x-axis, we obtain a graph whose equation is (3 4) 3 4.y x x
99. If the graph of ( ) 3 4f x x is reflected about the y-axis, we obtain a graph whose equation is ( ) 3( ) 4 3 4.y f x x x
100. If the graph of ( ) 3 4f x x is reflected about the origin, every point (x, y) will be replaced by the point (–x, –y). The equation for the graph will change from 3 4y x to
3( ) 4y x 3 4y x
3 4.y x
101. (a) To graph ( ) 3,y f x translate the graph of y = f(x), 3 units up.
(b) To graph ( 2),y f x translate the graph of y = f(x), 2 units to the right.
(c) To graph ( 3) 2,y f x translate the graph of y = f(x), 3 units to the left and 2 units down.
(d) To graph ( )y f x , keep the graph of
y = f(x) as it is where 0y and reflect the graph about the x-axis where y < 0.
102. No. For example suppose 2f x x and
2 .g x x Then
( )( ) ( ( )) (2 ) 2 2f g x f g x f x x
The domain and range of g are ( , ) ,
however, the domain of f is [2, ) . So,
2 2 0 1x x . Therefore, the domain of
f g is 1, . The domain of g, ( , ), is
not a subset of the domain of , 1, .f g
For Exercises 103–110, 2( ) 3 4f x x= − and 2( ) 3 4.g x x x= − −
103. 2 2
4 3 2 2
4 3 2
( )( ) ( ) ( )
(3 4)( 3 4)
3 9 12 4 12 16
3 9 16 12 16
fg x f x g xx x x
x x x x xx x x x
104. 2 2
( )(4) (4) (4)
(3 4 4) (4 3 4 4)(3 16 4) (16 3 4 4)(48 4) (16 12 4)44 0 44
f g f g
105. 2 2
( )( 4) ( 4) ( 4)
[3( 4) 4] [( 4) 3( 4) 4][3(16) 4] [16 3( 4) 4][48 4] [16 12 4]44 24 68
f g f g
106. 2 2
2 2
2 2
2
( )(2 ) (2 ) (2 )
[3(2 ) 4] [(2 ) 3(2 ) 4]
[3(4) 4] [4 3(2 ) 4]
(12 4) (4 6 4)
16 6 8
f g k f k g kk k kk k k
k k kk k
Chapter 2 Review Exercises 259
Copyright © 2017 Pearson Education, Inc.
107. 2
2
(3) 3 3 4 3 9 4(3)
(3) 9 3 3 43 3 3 427 4 23 23
9 9 4 4 4
f fg g
108.
2
2
3 1 4 3 1 4( 1)
1 3 1 41 3 1 43 4 1
undefined1 3 4 0
fg
109. The domain of (fg)(x) is the intersection of the domain of f(x) and the domain of g(x). Both have domain , , so the domain of
(fg)(x) is , .
110. 2 2
2
3 4 3 4( )
( 1)( 4)3 4
f x xxg x xx x
Because both f x and g x have domain
, , we are concerned about values of x
that make 0.g x Thus, the expression is
undefined if (x + 1)(x – 4) = 0, that is, if x = –1 or x = 4. Thus, the domain is the set of all real numbers except x = –1 and x = 4, or (– , –1) (–1, 4) (4, ).
111. 2 9f x x
( ) 2( ) 9 2 2 9f x h x h x h
( ) ( ) (2 2 9) (2 9)2 2 9 2 9 2
f x h f x x h xx h x h
Thus, ( ) ( ) 2
2.f x h f x h
h h
112. 2( ) 5 3f x x x 2
2 2( ) ( ) 5( ) 3
2 5 5 3
f x h x h x hx xh h x h
2 2 2
2 2 2
2
( ) ( )
( 2 5 5 3) ( 5 3)
2 5 5 3 5 3
2 5
f x h f xx xh h x h x x
x xh h x h x xxh h h
2( ) ( ) 2 5
(2 5)2 5
f x h f x xh h hh h
h x h x hh
For Exercises 113–118, 2( ) 2 and ( ) .f x x g x x
113. 2
( )( ) [ ( )] 2
2 2
g f x g f x g x
x x
114. 2 2( )( ) [ ( )] ( ) 2f g x f g x f x x
115. 2, so 3 3 2 1 1.f x x f
Therefore,
3 3 1g f g f g 21 1.
116. 22 , so 6 6 36.g x x g
Therefore, 6 6 36f g f g f
36 2 34 .
117. 1 1 1 2 3g f g f g g
Because 3 is not a real number, 1g f
is not defined.
118. To find the domain of ,f g we must consider the domain of g as well as the composed function, .f g Because
2 2f g x f g x x we need to
determine when 2 2 0.x Step 1: Find the values of x that satisfy
2 2 0.x 2 2 2x x
Step 2: The two numbers divide a number line into three regions.
Step 3 Choose a test value to see if it satisfies
the inequality, 2 2 0.x
Interval Test
Value Is 2 2 0x true or false?
, 2 2 22 2 0 ?2 0 True
2, 2 0 20 2 0 ?
2 0 False
2, 2 22 2 0 ?
2 0 True
The domain of f g is
, 2 2, .
260 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
119. 1 1 1 7 1 8f g f g
120. ( )(3) (3) (3) 9 9 0f g f g
121. ( )( 1) ( 1) ( 1) 3 2 6fg f g
122. (0) 5
(0) undefined(0) 0
f fg g
123. ( )( 2) [ ( 2)] (1) 2g f g f g
124. ( )(3) [ (3)] ( 2) 1f g f g f
125. ( )(2) [ (2)] (2) 1f g f g f
126. ( )(3) [ (3)] (4) 8g f g f g
127. Let x = number of yards. f(x) = 36x, where f(x) is the number of inches. g(x) = 1760x, where g(x) is the number of yards. Then
( )( ) ( ) 1760(36 )g f x g f x x 63,360 .x
There are 63,360x inches in x miles
128. Use the definition for the perimeter of a rectangle. P = length + width + length + width
( ) 2 2 6P x x x x x x This is a linear function.
129. If 343
( )V r r and if the radius is increased
by 3 inches, then the amount of volume gained is given by
3 34 43 3
( ) ( 3) ( ) ( 3) .gV r V r V r r r
130. (a) 2V r h If d is the diameter of its top, then h = d
and 2
.dr So,
32 2
2 4 4( ) ( ) ( ) .d d dV d d d
(b) 22 2S r rh
2
2 2 2
2 22 2 2
2 32 2 2
( ) 2 2 ( )d d d
d d d
S d d d
Chapter 2 Test
1. (a) The domain of 3f x x occurs
when 0.x In interval notation, this
correlates to the interval in D, 0, .
(b) The range of 3f x x is all real
numbers greater than or equal to 0. In interval notation, this correlates to the interval in D, 0, .
(c) The domain of 2 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(d) The range of 2 3f x x is all real
numbers greater than or equal to 3. In interval notation, this correlates to the interval in B, 3, .
(e) The domain of 3 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(f) The range of 3 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(g) The domain of 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(h) The range of 3f x x is all real
numbers greater than or equal to 0. In interval notation, this correlates to the interval in D, 0, .
(i) The domain of 2x y is 0x because when you square any value of y, the outcome will be nonnegative. In interval notation, this correlates to the interval in D, 0, .
(j) The range of 2x y is all real numbers. In interval notation, this correlates to the interval in C, , .
2. Consider the points 2,1 and 3, 4 .
4 1 3
3 ( 2) 5m
3. We label the points A 2,1 and B 3, 4 .
2 2
2 2
( , ) [3 ( 2)] (4 1)
5 3 25 9 34
d A B
Chapter 2 Test 261
Copyright © 2017 Pearson Education, Inc.
4. The midpoint has coordinates
2 3 1 4 1 5, , .
2 2 2 2
5. Use the point-slope form with 3
1 1 5( , ) ( 2,1) and .x y m
1 13535
( )
1 [ ( 2)]
1 ( 2) 5 1 3( 2)5 5 3 6 5 3 11
3 5 11 3 5 11
y y m x xy xy x y xy x y x
x y x y
6. Solve 3x – 5y = –11 for y.
3 115 5
3 5 115 3 11
x yy xy x
Therefore, the linear function is 3 115 5
( ) .f x x
7. (a) The center is at (0, 0) and the radius is 2, so the equation of the circle is
2 2 4x y .
(b) The center is at (1, 4) and the radius is 1, so the equation of the circle is
2 2( 1) ( 4) 1x y
8. 2 2 4 10 13 0x y x y Complete the square on x and y to write the equation in standard form:
2 2
2 2
2 2
2 2
4 10 13 0
4 10 13
4 4 10 25 13 4 25
2 5 16
x y x yx x y y
x x y y
x y
The circle has center (−2, 5) and radius 4.
9. (a) This is not the graph of a function because some vertical lines intersect it in more than one point. The domain of the relation is [0, 4]. The range is [–4, 4].
(b) This is the graph of a function because no vertical line intersects the graph in more than one point. The domain of the function is (– , –1) (–1, ). The
range is (– , 0) (0, ). As x is getting
larger on the intervals , 1 and
1, , the value of y is decreasing, so
the function is decreasing on these intervals. (The function is never increasing or constant.)
10. Point A has coordinates (5, –3). (a) The equation of a vertical line through A
is x = 5.
(b) The equation of a horizontal line through A is y = –3.
11. The slope of the graph of 3 2y x is –3.
(a) A line parallel to the graph of 3 2y x has a slope of –3.
Use the point-slope form with
1 1( , ) (2,3) and 3.x y m
1 1( )3 3( 2)3 3 6 3 9
y y m x xy xy x y x
(b) A line perpendicular to the graph of
3 2y x has a slope of 13
because
13
3 1.
13
713 3
3 ( 2)
3 3 2 3 9 2
3 7
y xy x y x
y x y x
12. (a) 2, (b) 0, 2
(c) , 0 (d) ,
(e) , (f) 1,
13. To graph 2 1f x x , we translate the
graph of ,y x 2 units to the right and 1 unit
down.
262 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
14. ( ) 1f x x
To get y = 0, we need 0 1 1x 1 0.x To get y = 1, we need
1 1 2x 0 1.x Follow this pattern to graph the step function.
15. 12
3 if 2( )
2 if 2x
f x x x
For values of x with x < –2, we graph the horizontal line y = 3. For values of x with
2,x we graph the line with a slope of 12
and a y-intercept of (0, 2). Two points on this line are (–2, 3) and (0, 2).
16. (a) Shift f(x), 2 units vertically upward.
(b) Shift f(x), 2 units horizontally to the left.
(c) Reflect f(x), across the x-axis.
(d) Reflect f(x), across the y-axis.
(e) Stretch f(x), vertically by a factor of 2.
17. Starting with ,y x we shift it to the left 2 units and stretch it vertically by a factor of 2. The graph is then reflected over the x-axis and then shifted down 3 units.
18. 2 23 2 3x y
(a) Replace y with –y to obtain 2 2 2 23 2( ) 3 3 2 3.x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the x-axis.
(b) Replace x with –x to obtain 2 2 2 23( ) 2 3 3 2 3.x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis.
(c) The graph is symmetric with respect to the x-axis and with respect to the y-axis, so it must also be symmetric with respect to the origin.
Chapter 2 Test 263
Copyright © 2017 Pearson Education, Inc.
19. 2( ) 2 3 2, ( ) 2 1f x x x g x x
(a)
2
2
2
( )( ) ( ) ( )
2 3 2 2 1
2 3 2 2 1
2 1
f g x f x g xx x x
x x xx x
(b) 2( ) 2 3 2
( )( ) 2 1
f f x x xxg g x x
(c) We must determine which values solve the equation 2 1 0.x
12
2 1 0 2 1x x x
Thus, 12
is excluded from the domain,
and the domain is 1 12 2
, , .
(d) 2( ) 2 3 2f x x x
2
2 2
2 2
2 3 2
2 2 3 3 2
2 4 2 3 3 2
f x h x h x hx xh h x h
x xh h x h
2 2
2
2 2
2
2
( ) ( )
(2 4 2 3 3 2)
(2 3 2)
2 4 2 3
3 2 2 3 2
4 2 3
f x h f xx xh h x h
x xx xh h x
h x xxh h h
2( ) ( ) 4 2 3
(4 2 3)
4 2 3
f x h f x xh h hh h
h x hh
x h
(e) 2
( )(1) (1) (1)
(2 1 3 1 2) ( 2 1 1)(2 1 3 1 2) ( 2 1 1)(2 3 2) ( 2 1)1 ( 1) 0
f g f g
(f) 2
( )(2) (2) (2)
(2 2 3 2 2) ( 2 2 1)(2 4 3 2 2) ( 2 2 1)(8 6 2) ( 4 1)4( 3) 12
fg f g
(g) 2 1 0 2 0 1g x x g
0 1 1. Therefore,
2
0 0
1 2 1 3 1 22 1 3 1 22 3 2 1
f g f gf
For exercises 20 and 21, 1f x x and
2 7.g x x
20. 2 7
(2 7) 1 2 6
f g f g x f xx x
The domain and range of g are ( , ) , while
the domain of f is [0, ) . We need to find the values of x which fit the domain of f: 2 6 0 3x x . So, the domain of f g
is [3, ) .
21. 1
2 1 7
g f g f x g x
x
The domain and range of g are ( , ) , while
the domain of f is [0, ) . We need to find the values of x which fit the domain of f:
1 0 1x x . So, the domain of g f
is [ 1, ) .
22. (a) C(x) = 3300 + 4.50x
(b) R(x) = 10.50x
(c) ( ) ( ) ( )10.50 (3300 4.50 )6.00 3300
P x R x C xx x
x
(d) ( ) 06.00 3300 0
6.00 3300550
P xx
xx
She must produce and sell 551 items before she earns a profit.
20 Copyright © 2017 Pearson Education, Inc
Chapter 2 Graphs and Functions Section 2.1 Rectangular Coordinates
and Graphs Classroom Example 1 (page 184)
(a) (transportation, $12,153)
(b) (health care, $4917)
Classroom Example 2 (page 186)
22( , ) 2 3 8 5
25 169 194
d P Q
Classroom Example 3 (page 186)
22
22
2 2
( , ) 5 0 1 2
25 9 34
( , ) 4 0 3 2
16 25 41
( , ) 4 5 3 1
81 4 85
d R S
d R T
d S T
The longest side has length 85
2 2 2?34 41 85
34 41 85
The triangle formed by the three points is not a right triangle.
Classroom Example 4 (page 187)
2 2
2 2
22
The distance between ( 2,5) and (0,3) is
2 0 5 3 4 4 8 2 2
The distance between (0,3) and (8, 5) is
8 0 5 3 64 64
128 8 2
The distance between ( 2,5) and (8, 5) is
2 8 5 5 100 100
200 10 2
P Q
Q R
P R
Because 2 2 8 2 10 2 , the points are collinear.
Classroom Example 5 (page 188)
(a) The coordinates of M are
7 ( 2) 5 13 9
, , 42 2 2
(b) Let (x, y) be the coordinates of Q. Use the midpoint formula to find the coordinates:
8 20, 4, 4
2 2
84 8 8 0
220
4 20 8 122
x y
xx x
yy y
The coordinates of Q are (0, 12).
Classroom Example 6 (page 188)
The year 2011 lies halfway between 2009 and 2013, so we must find the coordinates of the midpoint of the segment that has endpoints (2009, 124.0) and (2013, 137.4)
2009 2013 124.0 137.4,
2 2
2011, 130.7
M
The estimate of $130.7 billion is $0.1 billion more than the actual amount.
Classroom Example 7 (page 189)
Choose any real number for x, substitute the value in the equation and then solve for y. Note that additional answers are possible.
(a) x y = −2x + 5
−1 2( 1) 5 7y
0 2(0) 5 5y
3 2(3) 5 1y
Three ordered pairs that are solutions are (−1, 7), (0, 5), and (3, −1). Other answers are possible.
Section 2.2 Circles 21
Copyright © 2017 Pearson Education, Inc
(b) y 3 1x y
−9 3 39 1 8 2x
−2 3 32 1 1 1x
−1 3 31 1 0 0x
0 33 0 1 1 1x
7 3 37 1 8 2x
Ordered pairs that are solutions are (−2, −9), (−1, −2), (0, −1), (1, 0) and (2, 7). Other answers are possible.
(c) x 2 1y x
−2 2( 2) 1 3y
−1 2( 1) 1 0y
0 20 1 1y
1 21 1 0y
2 22 1 3y
Ordered pairs that are solutions are (−2, −3), (−1, 0), (0, 1), (1, 0) and (2, −3). Other answers are possible.
Classroom Example 8 (page 190)
(a) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:
52
0 2 52(0) 5 5x x
y y
Find a third point on the graph by letting x = −1 and solving for y: 2 1 5 7y .
The three points are 52
, 0 , (0, 5), and (−1, 7).
Note that (3, −1) is also on the graph.
(b) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:
33
3
0 1 1 1
0 1 0 1 1
x
y y y
Find a third point by letting x = 2 and solving
for y: 332 1 2 1 7y y y .
Find a fourth point by letting x = −2 and solving for y:
332 1 2 1 9y y y
The points to be plotted are (0, −1), (1, 0), (2, 7), and (−2, −9). Note that (−1, −2) is also on the graph.
(c) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:
2 2 2
2
0 1 1 1 1
0 1 1
x x x x
y
Find a third point by letting x = 2 and solving
for y: 22 1 3y . Find a fourth point
by letting x = −2 and solving for y:
22 1 3y
The points to be plotted are (−1, 0), (1, 0), (0, 1), (2, −3), and (−2, −3)
Section 2.2 Circles Classroom Example 1 (page 195)
(a) (h, k) = (1, −2) and r = 3
2 2 2
22 2
2 2
1 2 3
1 2 9
x h y k r
x y
x y
(b) (h, k) = (0, 0) and r = 2
2 2 2
2 2 2
2 2
0 0 2
4
x h y k r
x y
x y
22 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 2 (page 196)
(a) 2 21 2 9x y
This is a circle with center (1, −2) and radius 3.
(b) 2 2 4x y
This is a circle with center (0, 0) and radius 2.
Classroom Example 3 (page 197)
Complete the square twice, once for x and once for y:
2 2
2 2
2 2
4 8 44 0
4 4 8 16 44 4 16
2 4 64
x x y y
x x y y
x y
Because c = 64 and 64 > 0, the graph is a circle. The center is (−2, 4) and the radius is 8.
Classroom Example 4 (page 198)
2 22 2 2 6 45x y x y
Group the terms, factor out 2, and then complete the square:
2 21 92 2 3
4 4
1 945 2 2
4 4
x x y y
Factor and then divide both sides by 2:
2 2
2 2
1 32 2 50
2 2
1 325
2 2
x y
x y
Because c = 25 and 25 > 0, the graph is a circle. The
center is 1 3
,2 2
and the radius is 5.
Classroom Example 5 (page 198)
Complete the square twice, once for x and once for y:
2 2
2 2
2 2
6 2 12 0
6 9 2 1 12 9 1
3 1 2
x x y y
x x y y
x y
Because c = −2 and −2 < 0, the graph is nonexistent.
Classroom Example 6 (page 199)
Determine the equation for each circle and then substitute x = −3 and y = 4. Station A:
2 2 2
2 2 2
2 2
1 4 4
3 1 4 4 4
4 416 16
x y
Station B:
2 2 2
2 2
2 2
2 2
6 0 5
6 25
3 6 4 25
3 4 259 16 25
25 25
x y
x y
Station C:
22 2
2 2
2 2
2 2
5 2 10
5 2 100
3 5 4 2 100
8 6 10064 36 100
100 100
x y
x y
Because (−3, 4) satisfies all three equations, we can conclude that the epicenter is (−3, 4).
Section 2.3 Functions 23
Copyright © 2017 Pearson Education, Inc
Section 2.3 Functions Classroom Example 1 (page 204)
4,0 , 3,1 , 3,1M
M is a function because each distinct x value has exactly one y value.
2,3 , 3, 2 , 4,5 , 5, 4N
N is a function because each distinct x value has exactly one y value.
4,3 , 0,6 , 2,8 , 4, 3P
P is not a function because there are two y-values for x = −4.
Classroom Example 2 (page 205)
(a) Domain: {−4, −1, 1, 3} Range: {−2, 0, 2, 5} The relation is a function.
(b) Domain: {1, 2, 3} Range: {4, 5, 6, 7} The relation is not a function because 2 maps to 5 and 6.
(c) Domain: {−3, 0, 3, 5} Range: {5} The relation is a function.
Classroom Example 3 (page 206)
(a) Domain: {−2, 4}; range: {0, 3}
(b) Domain: ( , ) ; range: ( , )
(c) Domain: [−5, 5]; range: [−3, 3]
(d) Domain: ( , ) ; range: ( , 4]
Classroom Example 4 (page 207)
(a)
Not a function
(b)
Function
(c)
Not a function
(d)
Function
Classroom Example 5 (page 208)
(a) 2 5y x represents a function because y is
always found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers or ( , ) .
Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers, ( , ) .
24 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
(b) For any choice of x in the domain of 2 3y x , there is exactly one corresponding
value for y, so the equation defines a function. The function is defined for all values of x, so the domain is ( , ) . The square of any
number is always positive, so the range is [3, ) .
(c) For any choice of x in the domain of x y ,
there are two possible values for y. Thus, the equation does not define a function. The domain is [0, ) while the range is ( , ) .
(d) By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into y x corresponds to many values of y. The
ordered pairs (0, 2) (1, 1) (1, 0) (3, −1) and so on, all satisfy the inequality. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers, ( , ) .
(e) For 3
2y
x
, we see that y can be found by
dividing x + 2 into 3. This process produces one value of y for each value of x in the domain. The domain includes all real numbers except those that make the denominator equal to zero, namely x = −2. Therefore, the domain is ( , 2) ( 2, ) . Values of y can be
negative or positive, but never zero. Therefore the range is ( , 0) (0, ) .
Classroom Example 6 (page 210)
(a) 2( 3) ( 3) 6( 3) 4 13f
(b) 2( ) 6 4f r r r
(c) ( 2) 3( 2) 1 3 7g r r r
Classroom Example 7 (page 210)
(a) 2( 1) 2( 1) 9 7f
(b) ( 1) 6f
(c) ( 1) 5f
(d) ( 1) 0f
Classroom Example 8 (page 211)
(a) 2
2
2
( ) 2 3
( 5) ( 5) 2( 5) 3 12
( ) 2 3
f x x x
f
f t t t
(b) 2
2 3 6 23
2( ) 2
32 16
( 5) ( 5) 23 3
2( ) 2
3
x y y x
f x x
f
f t t
Section 2.4 Linear Functions 25
Copyright © 2017 Pearson Education, Inc
Classroom Example 9 (page 213)
The function is increasing on ( , 1), decreasing
on (–1, 1) and constant on (1, ).
Classroom Example 10 (page 213)
The example refers to the following figure.
(a) The water level is changing most rapidly from 0 to 25 hours.
(b) The water level starts to decrease after 50 hours.
(c) After 75 hours, there are 2000 gallons of water in the pool.
Section 2.4 Linear Functions Classroom Example 1 (page 220)
32
( ) 6f x x ; Use the intercepts to graph the
function. 32
(0) 0 6 6 :f y-intercept
3 32 2
0 6 6 4 :x x x x-intercept
Domain: ( , ), range: ( , )
Classroom Example 2 (page 220)
( ) 2f x is a constant function. Its graph is a
horizontal line with a y-intercept of 2.
Domain: ( , ), range: {2}
Classroom Example 3 (page 221)
x = 5 is a vertical line intersecting the x-axis at (5, 0).
Domain:{5}, range: ( , )
Classroom Example 4 (page 221)
3 4 0;x y Use the intercepts.
3 0 4 0 4 0 0 : -intercepty y y y
3 4 0 0 3 0 0 : -interceptx x x x
The graph has just one intercept. Choose an additional value, say 4, for x.
3 4 4 0 12 4 04 12 3
y yy y
Graph the line through (0, 0) and (4, −3).
Domain: ( , ), range: ( , )
26 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 5 (page 223)
(a) 4 ( 6) 10 5
2 2 4 2m
(b) 8 8 0
05 ( 3) 8
m
(c) 10 10 20
4 ( 4) 0m
the slope is
undefined.
Classroom Example 6 (page 224)
2x − 5y = 10 Solve the equation for y. 2 5 10
5 2 102
25
x yy x
y x
The slope is 2
,5
the coefficient of x.
Classroom Example 7 (page 224)
First locate the point (−2, −3). Because the slope is 43
, a change of 3 units horizontally (3 units to the
right) produces a change of 4 units vertically (4 units up). This gives a second point, (1, 1), which can be used to complete the graph.
Classroom Example 8 (page 225)
The average rate of change per year is 8658 9547 889
296.33 million2013 2010 3
The graph confirms that the line through the ordered pairs fall from left to right, and therefore has negative slope. Thus, the amount spent by the federal government on R&D for general science decreased by an average of $296.33 million (or $296,330,000) each year from 2010 to 2013.
Classroom Example 9 (page 226)
(a) ( ) 120 2400C x x
(b) ( ) 150R x x
(c) ( ) ( ) ( )150 (120 2400)30 2400
P x R x C xx x
x
(d) ( ) 0 30 2400 0 80P x x x
At least 81 items must be sold to make a profit.
Section 2.5 Equations of Lines and Linear Models
Classroom Example 1 (page 234)
( 5) 2( 3)5 2 6
2 1
y xy x
y x
Classroom Example 2 (page 234)
First find the slope: 3 ( 1) 4
4 5 9m
Now use either point for 1 1( , )x y : 49
3 [ ( 4)]9( 3) 4( 4)9 27 4 16
9 4 114 9 11
y xy xy x
y xx y
Classroom Example 3 (page 235)
Write the equation in slope-intercept form: 34
3 4 12 4 3 12 3x y y x y x
The slope is 34
, and the y-intercept is (0, −3).
Section 2.6 Graphs of Basic Functions 27
Copyright © 2017 Pearson Education, Inc
Classroom Example 4 (page 236)
First find the slope: 4 2 1
2 2 2m
Now, substitute 12
for m and the coordinates of one
of the points (say, (2, 2)) for x and y into the slope-intercept form y = mx + b, then solve for b:
12
2 2 3b b . The equation is
12
3y x .
Classroom Example 5 (page 236)
The example refers to the following figure:
(a) The line rises 5 units each time the x-value
increases by 2 units. So the slope is 52
. The
y-intercept is (0, 5), and the x-intercept is (−2, 0).
(b) An equation defining f is 52
( ) 5f x x .
Classroom Example 6 (page 238)
(a) Rewrite the equation 3 2 5x y in slope-
intercept form to find the slope: 3 52 2
3 2 5x y y x The slope is 32
.
The line parallel to the equation also has slope 32
. An equation of the line through (2, −4) that
is parallel to 3 2 5x y is 3 32 2
( 4) ( 2) 4 3y x y x
32
7y x or 3x − 2y = 14.
(b) The line perpendicular to the equation has
slope 23
. An equation of the line through
(2, −4) that is perpendicular to 3 2 5x y is 2 2 43 3 3
( 4) ( 2) 4y x y x 82
3 3y x or 2x +3y = −8.
Classroom Example 7 (page 240)
(a) First find the slope: 7703 6695
5043 1
m
Now use either point for 1 1( , )x y :
6695 504( 1)6695 504 504
504 6191
y xy x
y x
(b) The year 2015 is represented by x = 6. 504(6) 6191 9215y
According to the model, average tuition and fees for 4-year colleges in 2015 will be about $9215.
Classroom Example 8 (page 242)
Write the equation as an equivalent equation with 0 on one side.
3 2(5 ) 2 38x x x
3 2(5 ) 2 38 0x x x
Now graph the equation to find the x-intercept.
The solution set is {−4}.
Section 2.6 Graphs of Basic Functions Classroom Example 1 (page 249)
(a) The function is continuous over ( , 0) (0, )
(b) The function is continuous over its entire domain ( , ).
28 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 2 (page 252)
(a) Graph each interval of the domain separately. If x < 1, the graph of f(x) = 2x + 4 has an endpoint at (1, 6), which is not included as part of the graph. To find another point on this part of the graph, choose x = 0, so y = 4. Draw the ray starting at (1, 6) and extending through (0, 4). Graph the function for x ≥ 1, f(x) = 4 − x similarly. This part of the graph has an endpoint at (1, 3), which is included as part of the graph. Find another point, say (4, 0), and draw the ray starting at (1, 3) which extends through (4, 0).
(b) Graph each interval of the domain separately. If x ≤ 0, the graph of f(x) = −x − 2 has an endpoint at (0, −2), which is included as part of the graph. To find another point on this part of the graph, choose x = −2, so y = 0. Draw the ray starting at (0, −2) and extending through (−2, 0). Graph the function for x > 0,
2( ) 2f x x similarly. This part of the graph
has an endpoint at (0, −2), which is not included as part of the graph. Find another point, say (2, 2), and draw the curve starting at (0, −2) which extends through (2, 2). Note that the two endpoints coincide, so (0, −2) is included as part of the graph.
Classroom Example 3 (page 254)
Create a table of sample ordered pairs:
x −6 −3 32
0 32
3 6
13
2y x −4 −3 −3 −2 −2 −1 0
Classroom Example 4 (page 254)
For x in the interval (0, 2], y = 20. For x in (2, 3], y = 20 + 2 = 22. For x in (3, 4], y = 22 + 2 = 24. For x in (4, 5], y = 24 + 2 = 26. For x in (5, 6], y = 26 + 2 = 28.
Section 2.7 Graphing Techniques Classroom Example 1 (page 260)
Use this table of values for parts (a)−(c)
x 2( ) 2g x x 212
( )h x x 212
( )k x x
−2 8 2 1
−1 2 12
14
0 0 0 0
1 2 12
14
2 8 2 1
(a) 2( ) 2g x x
Section 2.7 Graphing Techniques 29
Copyright © 2017 Pearson Education, Inc
(b) 212
( )h x x
(c) 2
1( )
2k x x
Classroom Example 2 (page 262)
Use this table of values for parts (a) and (b)
x ( )g x x ( )h x x
−2 −2 2
−1 −1 1
0 0 0
1 −1 1
2 −2 2
(a) ( )g x x
(b) ( )h x x
Classroom Example 3 (page 263)
(a) x y
Replace x with –x to obtain x y . The result is not the same as the
original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain x y x y The
result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the x-axis only.
(b) 3y x
Replace x with –x to obtain 3y x
3y x . The result is the same as the
original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain 3y x . The result is not the same
as the original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to the y-axis only.
(c) 2 6x y
Replace x with –x to obtain 2( ) 6x y
2 6x y . The result is not the same as the
original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 2 ( ) 6x y
2 6x y . The result is not the same as the
original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is not symmetric with respect to either axis.
(d) 2 2 25x y
Replace x with –x to obtain 2 2( ) 25x y 2 2 25x y . The result is
the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2( ) 25x y 2 2 25x y . The result is
the same as the original equation, so the graph is symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to both axes. Note that the graph is a circle of radius 5, centered at the origin.
30 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 4 (page 265)
(a) 32y x
Replace x with –x and y with –y to obtain 3 3 3( ) 2( ) 2 2y x y x y x . The
result is the same as the original equation, so the graph is symmetric with respect to the origin.
(b) 22y x
Replace x with –x and y with –y to obtain 2 2 2( ) 2( ) 2 2y x y x y x .
The result is not the same as the original equation, so the graph is not symmetric with respect to the origin.
Classroom Example 5 (page 266)
(a) 5 3( ) 2 3g x x x x
Replace x with –x to obtain 5 3
5 3
5 3
( ) ( ) 2( ) 3( )
2 3
( 2 3 ) ( )
g x x x x
x x x
x x x g x
g(x) is an odd function.
(b) 2( ) 2 3h x x
Replace x with –x to obtain 2 2( ) 2( ) 3 2 3 ( )h x x x h x h(x) is
an even function.
(c) 2( ) 6 9k x x x
Replace x with –x to obtain 2
2( ) ( ) 6( ) 9
6 9 ( ) and ( )
k x x x
x x k x k x
k(x) is neither even nor odd.
Classroom Example 6 (page 267)
Compare a table of values for 2( )g x x with 2( ) 2f x x . The graph of f(x) is the same as the
graph of g(x) translated 2 units up.
x 2( )g x x 2( ) 2f x x
−2 4 6
−1 1 3
0 0 2
1 1 3
2 4 6
Classroom Example 7 (page 268)
Compare a table of values for 2( )g x x with 2( ) ( 4)f x x . The graph of f(x) is the same as the
graph of g(x) translated 4 units left.
x 2( )g x x 2( ) ( 4)f x x
−7 49 9
−6 36 4
−5 25 1
−4 16 0
−3 9 1
−2 4 4
−1 1 9
Classroom Example 8 (page 269)
(a) 2( ) ( 1) 4f x x
This is the graph of 2( )g x x , translated one
unit to the right, reflected across the x-axis, and then translated four units up.
x 2( )g x x 2( ) ( 1) 4f x x
−2 4 −5
−1 1 0
0 0 3
1 1 4
2 4 3
3 9 0
4 16 −5
(continued on next page)
Section 2.8 Function Operations and Composition 31
Copyright © 2017 Pearson Education, Inc
(continued)
(b) ( ) 2 3f x x
This is the graph of ( )g x x , translated three
units to the left, reflected across the x-axis, and then stretched vertically by a factor of two.
x ( )g x x ( ) 2 3f x x
−6 6 −6
−5 5 −4
−4 4 −2
−3 3 0
−2 2 −2
−1 1 −4
0 0 −6
(c) 12
( ) 2 3h x x
This is the graph of ( )g x x , translated two
units to the left, shrunk vertically by a factor of 2, and then translated 3 units down.
x ( )g x x 12
( ) 2 3h x x
−2 undefined −3
−1 undefined −2.5
0 0 12
2 3 2.3
2 2 1.4 −2
6 6 2.4 12
8 3 1.6
7 7 2.6 −1.5
Classroom Example 9 (page 270)
The graphs in the exercises are based on the following graph.
(a) ( ) ( ) 2g x f x
This is the graph of f(x) translated two units down.
(b) ( ) ( 2)h x f x
This is the graph of f(x) translated two units right.
32 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
(c) ( ) ( 1) 2k x f x
This is the graph of f(x) translated one unit left, and then translated two units up.
(d) ( ) ( )F x f x
This is the graph of f(x) reflected across the y-axis.
Section 2.8 Function Operations and Composition
Classroom Example 1 (page 278)
For parts (a)−(d), ( ) 3 4f x x and 2( ) 2 1g x x
(a) (0) 3(0) 4 4f and 2(0) 2(0) 1 1g , so
( )(0) 4 1 5f g
(b) (4) 3(4) 4 8f and 2(4) 2(4) 1 31g ,
so ( )(4) 8 31 23f g
(c) ( 2) 3( 2) 4 10f and 2( 2) 2( 2) 1 7g , so
( )( 2) ( 10)(7) 70fg
(d) (3) 3(3) 4 5f and 2(3) 2(3) 1 17g ,
so 5
(3)17
f
g
Classroom Example 2 (page 279)
For parts (a)−(e), 2( ) 3f x x x and ( ) 4 5g x x
(a) 2 2( )( ) ( 3 ) (4 5) 5f g x x x x x x
(b) 2 2( )( ) ( 3 ) (4 5) 7 5f g x x x x x x
(c) 2
3 2 2
3 2
( )( ) ( 3 )(4 5)
4 5 12 15
4 7 15
fg x x x x
x x x x
x x x
(d) 2 3
( )4 5
f x xx
g x
(e) The domains of f and g are both ( , ) . So,
the domains of f + g, f − g, and fg are the intersection of the domains of f and g,
( , ) . The domain of fg
includes those
real numbers in the intersection of the domains of f and g for which ( ) 4 5 0g x x
54
x . So the domain of fg
is
5 54 4
, , .
Classroom Example 3 (page 280)
(a)
From the figure, we have f(1) = 3 and
g(1) = 1, so (f + g)(1) = 3 + 1 = 4. f(0) = 4 and g(0) = 0, so (f − g)(0) = 4 − 0 = 4. f(−1) = 3 and g(−1) = 1, so (f g)(−1) = (3)(1) = 3
f(−2) = 0 and g(−2) = 2, so 02
( 2) 0fg
.
(b) x f(x) g(x)
−2 −5 0
−1 −3 2
0 −1 4
1 1 6
From the table, we have f(1) = 1 and g(1) = 6, so (f + g)(1) = 1 + 6 = 7. f(0) = −1 and g(0) = 4, so (f − g)(1) = −1 − 4 = −5. f(−1) = −3 and g(−1) = 2, so (f g)(−1) = (−3)(2) = −6 f(−2) = −5 and g(−2) = 0, so
50
( 2)fg
fg
is undefined.
Section 2.8 Function Operations and Composition 33
Copyright © 2017 Pearson Education, Inc
(c) ( ) 3 4, ( )f x x g x x
From the formulas, we have (1) 3(1) 4 7f and (1) 1 1g , so
(f + g)(1) = 7 + (−1) = 6. (0) 3(0) 4 4f and (0) 0 0g , so
(f − g)(1) = 4 − 0 = 4. ( 1) 3( 1) 4 1f and ( 1) 1 1g ,
so (fg)(−1) =(1)(−1) = −1. ( 2) 3( 2) 4 2f and
( 2) 2 2g , so 2
( 2) 1.2
f
g
Classroom Example 4 (page 281)
Step 1: Find ( )f x h : 2
2 2
2 2
( ) 3( ) 2( ) 4
3( 2 ) 2 2 4
3 6 3 2 2 4
f x h x h x h
x xh h x h
x xh h x h
Step 2: Find ( ) ( )f x h f x :
2 2 2
2
( ) ( )
(3 6 3 2 2 4) (3 2 4)
6 3 2
f x h f x
x xh h x h x x
xh h h
Step 3: Find the difference quotient: 2( ) ( ) 6 3 2
6 3 2f x h f x xh h h
x hh h
Classroom Example 5 (page 283)
For parts (a) and (b), ( ) 4f x x and 2
( )g xx
(a) First find g(2): 2
(2) 12
g . Now find
( )(2) ( (2)) (1) 1 4 5f g f g f
(b) First find f(5): (5) 5 4 9 3f . Now
find 23
( )(5) ( (5)) (3)g f g f g
The screens show how a graphing calculator
evaluates the expressions in this classroom example.
Classroom Example 6 (page 283)
For parts (a) and (b), ( ) 1f x x and
( ) 2 5g x x
(a) ( )( ) ( ( )) (2 5) 1 2 4f g x f g x x x
The domain and range of g are both ( , ) .
However, the domain of f is [1, ) . Therefore,
( )g x must be greater than or equal to 1:
2 5 1 2x x . So, the domain of f g
is [ 2, ) .
(b) ( )( ) ( ( )) 2 1 5g f x g f x x
The domain of f is [1, ) , while the range of f
is [0, ) . The domain of g is ( , ) .
Therefore, the domain of ( )g f is restricted
to that portion of the domain of g that intersects with the domain of f, that is [1, ) .
Classroom Example 7 (page 284)
For parts (a) and (b), 5
( )4
f xx
and 2
( )g xx
(a) 5 5
( )( ) ( ( ))2 4 2 4
xf g x f g x
x x
The domain and range of g are both all real numbers except 0. The domain of f is all real numbers except −4. Therefore, the expression for ( )g x cannot equal −4. So,
2 14
2x
x . So, the domain of f g
is the set of all real numbers except for 12
,
and 0. This is written
1 12 2
, , 0 0,
34 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
(b) 2 2 8
( )( ) ( ( ))5 ( 4) 5
xg f x g f x
x
The domain of f is all real numbers except −4, while the range of f is all real numbers except 0. The domain and range of g are both all real numbers except 0, which is not in the range of f. So, the domain of g f is the set of all real
numbers except for −4. This is written , 4 4,
Classroom Example 8 (page 285)
( ) 2 5f x x and 2( ) 3g x x x 2
2
2
2 2
2
( )( ) (2 5) 3(2 5) (2 5)
3(4 20 25) 2 5
12 58 70
( )( ) (3 ) 2(3 ) 5
6 2 5
g f x g x x x
x x x
x x
f g x f x x x x
x x
In general, 2 212 58 70 6 2 5,x x x x so
( )( ) ( )( )g f x f g x .
Classroom Example 9 (page 286)
2( )( ) 4(3 2) 5(3 2) 8f g x x x
Answers may vary. Sample answer: 2( ) 4 5 8f x x x and ( ) 3 2g x x .
Copyright © 2017 Pearson Education, Inc. 170
Chapter 2 GRAPHS AND FUNCTIONS
Section 2.1 Rectangular Coordinates and Graphs
1. The point (–1, 3) lies in quadrant II in the rectangular coordinate system.
2. The point (4, 6) lies on the graph of the equation y = 3x – 6. Find the y-value by letting x = 4 and solving for y.
3 4 6 12 6 6y
3. Any point that lies on the x-axis has y-coordinate equal to 0.
4. The y-intercept of the graph of y = –2x + 6 is (0, 6).
5. The x-intercept of the graph of 2x + 5y = 10 is (5, 0). Find the x-intercept by letting y = 0 and solving for x.
2 5 0 10 2 10 5x x x
6. The distance from the origin to the point (–3, 4) is 5. Using the distance formula, we have
2 2
2 2
( , ) ( 3 0) (4 0)
3 4 9 16 25 5
d P Q
7. True
8. True
9. False. The midpoint of the segment joining (0, 0) and (4, 4) is
4 0 4 0 4 4, , 2, 2 .
2 2 2 2
10. False. The distance between the point (0, 0) and (4, 4) is
2 2 2 2( , ) (4 0) (4 0) 4 4
16 16 32 4 2
d P Q
11. Any three of the following: 2, 5 , 1,7 , 3, 9 , 5, 17 , 6, 21
12. Any three of the following: 3,3 , 5, 21 , 8,18 , 4,6 , 0, 6
13. Any three of the following: (1999, 35), (2001, 29), (2003, 22), (2005, 23), (2007, 20), (2009, 20)
14. Any three of the following:
2002,86.8 , 2004, 89.8 , 2006, 90.7 ,
2008, 97.4 , 2010, 106.5 , 2012,111.4 ,
2014, 111.5
15. P(–5, –6), Q(7, –1)
(a) 2 2
2 2
( , ) [7 – (–5)] [ 1 – (–6)]
12 5 169 13
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
–5 7 6 ( 1) 2 7, ,
2 2 2 27
1, .2
Section 2.1 Rectangular Coordinates and Graphs 171
Copyright © 2017 Pearson Education, Inc.
16. P(–4, 3), Q(2, –5)
(a) 2 2
2 2
( , ) [2 – (– 4)] (–5 – 3)
6 (–8) 100 10
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 4 2 3 (–5) 2 2, ,
2 2 2 21, 1 .
17. (8, 2), (3, 5)P Q
(a)
2 2
2 2
( , ) (3 – 8) (5 – 2)
5 3
25 9 34
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
8 3 2 5 11 7, ,
2 2 2 2
.
18. P (−8, 4), Q (3, −5)
(a) 2 2
2 2
( , ) 3 – 8 5 4
11 (–9) 121 81
202
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
–8 3 4 (–5) 5 1, , .
2 2 2 2
19. P(–6, –5), Q(6, 10)
(a) 2 2
2 2
( , ) [6 – (– 6)] [10 – (–5)]
12 15 144 225
369 3 41
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 6 6 –5 10 0 5 5, , 0, .
2 2 2 2 2
20. P(6, –2), Q(4, 6)
(a) 2 2
2 2
( , ) (4 – 6) [6 – (–2)]
(–2) 8
4 64 68 2 17
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
6 4 2 6 10 4, , 5, 2
2 2 2 2
21. 3 2, 4 5 ,P 2, – 5Q
(a)
2 2
2 2
( , )
2 – 3 2 – 5 – 4 5
–2 2 –5 5
8 125 133
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
3 2 2 4 5 (– 5),
2 2
4 2 3 5 3 5, 2 2, .
2 2 2
22. – 7, 8 3 , 5 7, – 3P Q
(a) 2 2
2 2
( , )
[5 7 – (– 7)] (– 3 – 8 3)
(6 7) (–9 3) 252 243
495 3 55
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 7 5 7 8 3 (– 3),
2 2
4 7 7 3 7 3, 2 7, .
2 2 2
23. Label the points A(–6, –4), B(0, –2), and C(–10, 8). Use the distance formula to find the length of each side of the triangle.
2 2
2 2
( , ) [0 – (– 6)] [–2 – (– 4)]
6 2 36 4 40
d A B
2 2
2 2
( , ) (–10 – 0) [8 – (–2)]
( 10) 10 100 100
200
d B C
2 2
2 2
( , ) [–10 – (– 6)] [8 – (– 4)]
(– 4) 12 16 144 160
d A C
Because 2 2 240 160 200 ,
triangle ABC is a right triangle.
172 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
24. Label the points A(–2, –8), B(0, –4), and C(–4, –7). Use the distance formula to find the length of each side of the triangle.
2 2
2 2
( , ) [0 – (–2)] [– 4 – (–8)]
2 4 4 16 20
d A B
2 2
2 2
( , ) (– 4 – 0) [–7 – (– 4)]
(– 4) (–3) 16 9
25 5
d B C
2 2
2 2
( , ) [– 4 – (–2)] [–7 – (–8)]
(–2) 1 4 1 5
d A C
Because 2 2 2( 5) ( 20) 5 20 25 5 , triangle ABC is a right triangle.
25. Label the points A(–4, 1), B(1, 4), and C(–6, –1).
2 2
2 2
( , ) [1 – (– 4)] (4 – 1)
5 3 25 9 34
d A B
2 2
2 2
( , ) (– 6 –1) (–1 – 4)
(–7) (–5) 49 25 74
d B C
2 2
2 2
( , ) [– 6 – (– 4)] (–1 –1)
(–2) (–2) 4 4 8
d A C
Because 2 2 2( 8) ( 34) ( 74) because
8 34 42 74, triangle ABC is not a right triangle.
26. Label the points A(–2, –5), B(1, 7), and C(3, 15).
2 2
2 2
( , ) [1 ( 2)] [7 ( 5)]
3 12 9 144 153
d A B
2 2
2 2
( , ) (3 1) (15 7)
2 8 4 64 68
d B C
2 2
2 2
( , ) [3 ( 2)] [15 ( 5)]
5 20 25 400 425
d A C
Because 22 2( 68) ( 153) 425 because
68 153 221 425 , triangle ABC is not a right triangle.
27. Label the points A(–4, 3), B(2, 5), and C(–1, –6).
2 2
2 2
( , ) 2 – –4 5 3
6 2 36 4 40
d A B
2 2
2 2
( , ) 1 2 6 5
(– 3) (–11)
9 121 130
d B C
2 2
22
( , ) –1 – –4 6 3
3 9 9 81 90
d A C
Because 2 2 240 90 130 , triangle
ABC is a right triangle.
28. Label the points A(–7, 4), B(6, –2), and C(0, –15).
2 2
22
( , ) 6 – –7 2 4
13 6
169 36 205
d A B
22
2 2
( , ) 0 6 –15 – –2
– 6 –13
36 169 205
d B C
2 2
22
( , ) 0 – –7 15 4
7 19 49 361 410
d A C
Because 2 2 2205 205 410 ,
triangle ABC is a right triangle.
29. Label the given points A(0, –7), B(–3, 5), and C(2, –15). Find the distance between each pair of points.
22
2 2
( , ) 3 0 5 – –7
–3 12 9 144
153 3 17
d A B
2 2
22
( , ) 2 – 3 –15 – 5
5 –20 25 400
425 5 17
d B C
22
22
( , ) 2 0 15 – 7
2 –8 68 2 17
d A C
Because ( , ) ( , ) ( , )d A B d A C d B C or
3 17 2 17 5 17, the points are collinear.
Section 2.1 Rectangular Coordinates and Graphs 173
Copyright © 2017 Pearson Education, Inc.
30. Label the points A(–1, 4), B(–2, –1), and C(1, 14). Apply the distance formula to each pair of points.
2 2
2 2
( , ) –2 – –1 –1 – 4
–1 –5 26
d A B
2 2
2 2
( , ) 1 – –2 14 – –1
3 15 234 3 26
d B C
2 2
2 2
( , ) 1 – –1 14 – 4
2 10 104 2 26
d A C
Because 26 2 26 3 26 , the points are collinear.
31. Label the points A(0, 9), B(–3, –7), and C(2, 19).
2 2
2 2
( , ) (–3 – 0) (–7 – 9)
(–3) (–16) 9 256
265 16.279
d A B
2 2
2 2
( , ) 2 – –3 19 – –7
5 26 25 676
701 26.476
d B C
2 2
2 2
( , ) 2 – 0 19 – 9
2 10 4 100
104 10.198
d A C
Because ( , ) ( , ) ( , )d A B d A C d B C
or 265 104 70116.279 10.198 26.476,
26.477 26.476,
the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)
32. Label the points A(–1, –3), B(–5, 12), and C(1, –11).
2 2
2 2
( , ) –5 – –1 12 – –3
– 4 15 16 225
241 15.5242
d A B
2 2
22
( , ) 1 – –5 –11 –12
6 –23 36 529
565 23.7697
d B C
2 2
22
( , ) 1 – –1 –11 – –3
2 –8 4 64
68 8.2462
d A C
Because d(A, B) + d(A, C) d(B, C)
or 241 68 56515.5242 8.2462 23.7697
23.7704 23.7697,
the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)
33. Label the points A(–7, 4), B(6,–2), and C(–1,1).
2 2
22
( , ) 6 – –7 2 4
13 6 169 36
205 14.3178
d A B
22
2 2
( , ) 1 6 1 –2
7 3 49 9
58 7.6158
d B C
2 2
22
( , ) 1 – –7 1 4
6 –3 36 9
45 6.7082
d A C
Because d(B, C) + d(A, C) d(A, B) or
58 45 2057.6158 6.7082 14.3178
14.3240 14.3178,
the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)
34. Label the given points A(–4, 3), B(2, 5), and C(–1, 4). Find the distance between each pair of points.
2 2 2 2( , ) 2 – –4 5 3 6 2
36 4 40 2 10
d A B
2 2
2 2
( , ) ( 1 2) (4 5)
3 (–1) 9 1 10
d B C
2 2
2 2
( , ) 1 – 4 4 3
3 1 9 1 10
d A C
Because ( , ) ( , ) ( , )d B C d A C d A B or
10 10 2 10, the points are collinear.
174 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
35. Midpoint (5, 8), endpoint (13, 10) 13 10
5 and 82 2
13 10 and 10 16
–3 and 6.
x y
x yx y
The other endpoint has coordinates (–3, 6).
36. Midpoint (–7, 6), endpoint (–9, 9) –9 9
–7 and 62 2
–9 –14 and 9 12
–5 and 3.
x y
x yx y
The other endpoint has coordinates (–5, 3).
37. Midpoint (12, 6), endpoint (19, 16) 19 16
12 and 62 2
19 24 and 16 12
5 and – 4.
x y
x yx y
The other endpoint has coordinates (5, –4).
38. Midpoint (–9, 8), endpoint (–16, 9) –16 9
–9 and 82 2
–16 –18 and 9 16
–2 and 7
x y
x yx y
The other endpoint has coordinates (–2, 7).
39. Midpoint (a, b), endpoint (p, q)
and2 2
2 and 2
2 and 2
p x q ya b
p x a q y bx a p y b q
The other endpoint has coordinates (2 , 2 )a p b q .
40. Midpoint 6 , 6a b , endpoint 3 , 5a b
3 56 and 6
2 23 12 and 5 12
9 and 7
a x b ya b
a x a b y bx a y b
The other endpoint has coordinates (9a, 7b).
41. The endpoints of the segment are (1990, 21.3) and (2012, 30.1).
1990 2012 21.3 30.9,
2 22001, 26.1
M
The estimate is 26.1%. This is very close to the actual figure of 26.2%.
42. The endpoints are (2006, 7505) and (2012, 3335)
2006 2012 7505 3335,
2 22009, 5420
M
According to the model, the average national advertising revenue in 2009 was $5420 million. This is higher than the actual value of $4424 million.
43. The points to use are (2011, 23021) and (2013, 23834). Their midpoint is
2011 2013 23,021 23,834,
2 2(2012, 23427.5).
In 2012, the poverty level cutoff was approximately $23,428.
44. (a) To estimate the enrollment for 2003, use the points (2000, 11,753) and (2006, 13,180)
2000 2006 11,753 13,180,
2 22003, 12466.5
M
The enrollment for 2003 was about 12,466.5 thousand.
(b) To estimate the enrollment for 2009, use the points (2006, 13,180) and (2012, 14,880)
2006 2012 13,180 14,880,
2 22009, 14030
M
The enrollment for 2009 was about 14,030 thousand.
45. The midpoint M has coordinates
1 2 1 2, .2 2
x x y y
2 21 2 1 2
1 1
2 21 2 1 1 2 1
2 22 1 2 1
2 22 1 2 1
2 22 1 2 1
2 212 1 2 12
( , )
– –2 2
2 2– –
2 2 2 2
2 2
4 4
4
d P M
x x y yx y
x x x y y y
x x y y
x x y y
x x y y
x x y y
(continued on next page)
Section 2.1 Rectangular Coordinates and Graphs 175
Copyright © 2017 Pearson Education, Inc.
(continued)
2 21 2 1 2
2 2
2 22 1 2 2 1 2
2 22 1 2 1
2 22 1 2 1
2 22 1 2 1
2 212 1 2 12
( , )
2 2
2 2
2 2 2 2
2 2
4 4
4
d M Q
x x y yx y
x x x y y y
x x y y
x x y y
x x y y
x x y y
2 22 1 2 1( , )d P Q x x y y
Because
2 212 1 2 12
2 212 1 2 12
2 22 1 2 1 ,
x x y y
x x y y
x x y y
this shows ( , ) ( , ) ( , )d P M d M Q d P Q and
( , ) ( , ).d P M d M Q
46. The distance formula, 2 2
2 1 2 1( – ) ( – ) ,d x x y y can be written
as 2 2 1/ 22 1 2 1[( – ) ( – ) ] .d x x y y
In exercises 47−58, other ordered pairs are possible.
47. (a) x y 0 −2 y-intercept:
12
00 2 2
xy
4 0 x-intercept:
1212
00 2
2 4
yxx x
2 −1 additional point
(b)
48. (a) x y 0 2 y-intercept:
0
10 2 2
2
x
y
4 0 x-intercept: 0
10 2
21
2 42
y
x
x x
2 1 additional point
(b)
49. (a) x y 0 5
3 y-intercept:
53
02 0 3 5
3 5
xy
y y
52
0 x-intercept:
52
02 3 0 5
2 5
yxx x
4 −1 additional point
(b)
176 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
50. (a) x y 0 −3 y-intercept:
0
3 0 2 62 6 3
xy
y y
2 0 x-intercept:
0
3 2 0 63 6 2
yxx x
4 3 additional point
(b)
51. (a) x y 0 0 x- and y-intercept:
20 0
1 1 additional point
−2 4 additional point
(b)
52. (a) x y 0 2 y-intercept:
2
0
0 20 2 2
xyy y
−1 3 additional point 2 6 additional point
no x-intercept: 2
20 0 2
2 2
y xx x
(b)
53. (a) x y 3 0 x-intercept:
0
0 30 3 3
yx
x x
4 1 additional point 7 2 additional point
no y-intercept:
0 0 3 3x y y
(b)
54. (a) x y 0 −3 y-intercept:
0
0 30 3 3
xyy y
4 −1 additional point
9 0 x-intercept: 0
0 3
3 9
yxx x
(b)
Section 2.1 Rectangular Coordinates and Graphs 177
Copyright © 2017 Pearson Education, Inc.
55. (a) x y 0 2 y-intercept:
00 2
2 2
xyy y
2 0 x-intercept: 0
0 20 2 2
yxx x
−2 4 additional point
4 2 additional point
(b)
56. (a) x y −2 −2 additional point −4 0 x-intercept:
00 4
0 40 4 4
yx
xx x
0 −4 y-intercept: 0
0 4
4 4
xyy y
(b)
57. (a) x y 0 0 x- and y-intercept:
30 0 −1 −1 additional point 2 8 additional point
(b)
58. (a) x y 0 0 x- and y-intercept:
30 0 1 −1 additional point2 −8 additional point
(b)
59. Points on the x-axis have y-coordinates equal to 0. The point on the x-axis will have the same x-coordinate as point (4, 3). Therefore, the line will intersect the x-axis at (4, 0).
60. Points on the y-axis have x-coordinates equal to 0. The point on the y-axis will have the same y-coordinate as point (4, 3). Therefore, the line will intersect the y-axis at (0, 3).
61. Because (a, b) is in the second quadrant, a is negative and b is positive. Therefore, (a, – b) will have a negative x–coordinate and a negative y-coordinate and will lie in quadrant III. (–a, b) will have a positive x-coordinate and a positive y-coordinateand will lie in quadrant I. (–a, – b) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. (b, a) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV.
178 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
62. Label the points ( 2, 2),A (13,10),B
(21, 5),C and (6, 13).D To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.
Use the distance formula to find the length of
each side.
2 2
2 2
2 2
22
( , ) 13 2 10 2
15 8 225 64
289 17
( , ) 21 13 5 10
8 15 64 225
289 17
d A B
d B C
22
2 2
( , ) 6 21 13 5
15 8
225 64 289 17
d C D
22
2 2
( , ) 2 6 2 13
8 15
64 225 289 17
d D A
Because all sides have equal length, the four points form a rhombus.
63. To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.
Use the distance formula to find the length of
each side.
2 2
2 2
( , ) 5 1 2 1
4 1 16 1 17
d A B
2 2
2 2
( , ) 3 5 4 2
2 2 4 4 8
d B C
2 2
2 2
( , ) 1 3 3 4
4 1
16 1 17
d C D
2 2
22
( , ) 1 1 1 3
2 2 4 4 8
d D A
Because d(A, B) = d(C, D) and d(B, C) = d(D, A), the points are the vertices of a parallelogram. Because d(A, B) ≠ d(B, C), the points are not the vertices of a rhombus.
64. For the points A(4, 5) and D(10, 14), the difference of the x-coordinates is 10 – 4 = 6 and the difference of the y-coordinates is 14 – 5 = 9. Dividing these differences by 3, we obtain 2 and 3, respectively. Adding 2 and 3 to the x and y coordinates of point A, respectively, we obtain B(4 + 2, 5 + 3) or B(6, 8).
Adding 2 and 3 to the x- and y- coordinates of point B, respectively, we obtain C(6 + 2, 8 + 3) or C(8, 11). The desired points are B(6, 8) and C(8, 11).
We check these by showing that d(A, B) = d(B, C) = d(C, D) and that d(A, D) = d(A, B) + d(B, C) + d(C, D).
2 2
2 2
2 2
2 2
( , ) 6 4 8 5
2 3 4 9 13
( , ) 8 6 11 8
2 3 4 9 13
d A B
d B C
2 2
2 2
2 2
2 2
( , ) 10 8 14 11
2 3 4 9 13
( , ) 10 4 14 5
6 9 36 81
117 9(13) 3 13
d C D
d A D
d(A, B), d(B, C), and d(C, D) all have the same measure and d(A, D) = d(A, B) + d(B, C) + d(C, D) Because
3 13 13 13 13.
Section 2.2 Circles 179
Copyright © 2017 Pearson Education, Inc.
Section 2.2 Circles
1. The circle with equation 2 2 49x y has center with coordinates (0, 0) and radius equal to 7.
2. The circle with center (3, 6) and radius 4 has
equation 23 6 16.x y
3. The graph of 2 24 7 9x y has center
with coordinates (4, –7).
4. The graph of 22 5 9x y has center with
coordinates (0, 5).
5. This circle has center (3, 2) and radius 5. This is graph B.
6. This circle has center (3, –2) and radius 5. This is graph C.
7. This circle has center (–3, 2) and radius 5. This is graph D.
8. This circle has center (–3, –2) and radius 5. This is graph A.
9. The graph of 2 2 0x y has center (0, 0) and radius 0. This is the point (0, 0). Therefore, there is one point on the graph.
10. 100 is not a real number, so there are no
points on the graph of 2 2 100.x y
11. (a) Center (0, 0), radius 6
2 2
2 2 2 2 2
0 0 6
0 0 6 36
x y
x y x y
(b)
12. (a) Center (0, 0), radius 9
2 2
2 2 2 2 2
0 0 9
0 0 9 81
x y
x y x y
(b)
13. (a) Center (2, 0), radius 6
2 2
2 2 2
2 2
2 0 6
2 0 6
( – 2) 36
x y
x yx y
(b)
14. (a) Center (3, 0), radius 3
2 2
2 2
3 0 3
3 9
x yx y
(b)
15. (a) Center (0, 4), radius 4
2 2
22
0 4 4
4 16
x yx y
(b)
180 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
16. (a) Center (0, –3), radius 7
22
22 2
2 2
0 3 7
0 3 7
( 3) 49
x y
x yx y
(b)
17. (a) Center (–2, 5), radius 4
2 2
2 2 2
2 2
2 5 4
[ – (–2)] ( – 5) 4
( 2) ( – 5) 16
x y
x yx y
(b)
18. (a) Center (4, 3), radius 5
2 2
2 2 2
2 2
4 3 5
4 3 5
4 3 25
x y
x yx y
(b)
19. (a) Center (5, –4), radius 7
22
2 2 2
2 2
5 4 7
( – 5) [ – (– 4)] 7
( – 5) ( 4) 49
x y
x yx y
(b)
20. (a) Center (–3, –2), radius 6
2 2
2 2 2
2 2
3 2 6
3 2 6
( 3) ( 2) 36
x y
x yx y
(b)
21. (a) Center 2, 2 , radius 2
2 2
2 2
2 2 2
2 2 2
x y
x y
(b)
22. (a) Center 3, 3 , radius 3
2 2
2 2 2
2 2
3 3 3
3 3 3
3 3 3
x y
x y
x y
Section 2.2 Circles 181
Copyright © 2017 Pearson Education, Inc.
(b)
23. (a) The center of the circle is located at the midpoint of the diameter determined by the points (1, 1) and (5, 1). Using the midpoint formula, we have
1 5 1 1, 3,1
2 2C
. The radius is
one-half the length of the diameter:
2 215 1 1 1 2
2r
The equation of the circle is
2 23 1 4x y
(b) Expand 2 23 1 4x y to find the
equation of the circle in general form:
2 2
2 2
2 2
3 1 46 9 2 1 4
6 2 6 0
x yx x y y
x y x y
24. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−1, 1) and (−1, −5).
Using the midpoint formula, we have
1 ( 1) 1 ( 5), 1, 2 .
2 2C
The radius is one-half the length of the diameter:
2 211 1 5 1 3
2r
The equation of the circle is
2 21 2 9x y
(b) Expand 2 21 2 9x y to find the
equation of the circle in general form:
2 2
2 2
2 2
1 2 92 1 4 4 9
2 4 4 0
x yx x y y
x y x y
25. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−2, 4) and (−2, 0). Using the midpoint formula, we have
2 ( 2) 4 0, 2, 2 .
2 2C
The radius is one-half the length of the diameter:
2 212 2 4 0 2
2r
The equation of the circle is
2 22 2 4x y
(b) Expand 2 22 2 4x y to find the
equation of the circle in general form:
2 2
2 2
2 2
2 2 44 4 4 4 4
4 4 4 0
x yx x y y
x y x y
26. (a) The center of the circle is located at the midpoint of the diameter determined by the points (0, −3) and (6, −3). Using the midpoint formula, we have
0 6 3 ( 3), 3, 3
2 2C
.
The radius is one-half the length of the diameter:
2216 0 3 3 3
2r
The equation of the circle is
2 23 3 9x y
(b) Expand 2 23 3 9x y to find the
equation of the circle in general form:
2 2
2 2
2 2
3 3 96 9 6 9 9
6 6 9 0
x yx x y y
x y x y
27. 2 2 6 8 9 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
6 8 –9
6 9 8 16 –9 9 16
3 4 16
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (–3, –4) and radius 4.
182 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
28. 2 2 8 – 6 16 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
8 – 6 16
8 16 – 6 9 –16 16 9
4 – 3 9
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (–4, 3) and radius 3.
29. 2 2 4 12 4x y x y Complete the square on x and y separately.
2 2
2 2
2 2
– 4 12 – 4
– 4 4 12 36 – 4 4 36
– 2 6 36
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (2, –6) and radius 6.
30. 2 2 –12 10 –25x y x y Complete the square on x and y separately.
2 2
2 2
2 2
–12 10 –25
–12 36 10 25
– 25 36 25
– 6 5 36
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (6, –5) and radius 6.
31. 2 24 4 4 –16 – 19 0x y x y Complete the square on x and y separately.
2 2
2 214
14
4 4 – 4 19
4 4 – 4 4
19 4 4 4
x x y y
x x y y
2 212
2 212
4 4 – 2 36
– 2 9
x y
x y
Yes, it is a circle with center 12
, 2 and
radius 3.
32. 2 29 9 12 – 18 – 23 0x y x y Complete the square on x and y separately.
2 243
2 24 43 9
49
9 9 – 2 23
9 9 – 2 1
23 9 9 1
x x y y
x x y y
2 223
2 223
9 9 –1 36
–1 4
x y
x y
Yes, it is a circle with center 23
, 1 and
radius 2.
33. 2 2 2 – 6 14 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
2 – 6 –14
2 1 – 6 9 –14 1 9
1 – 3 – 4
x x y y
x x y y
x y
The graph is nonexistent.
34. 2 2 4 – 8 32 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
4 – 8 –32
4 4 – 8 16
– 32 4 16
2 – 4 –12
x x y y
x x y y
x y
The graph is nonexistent.
35. 2 2 6 6 18 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
6 6 18
6 9 6 9 18 9 9
3 3 0
x x y y
x x y y
x y
The graph is the point (3, 3).
36. 2 2 4 4 8 0x y x y Complete the square on x and y separately.
2 2
2 2
2 2
4 4 8
4 4 4 4 8 4 4
2 2 0
x x y y
x x y y
x y
The graph is the point (−2, −2).
37. 2 29 9 6 6 23 0x y x y Complete the square on x and y separately.
2 2
2 22 23 3
22 2 232 1 2 1 1 13 9 3 9 9 9 9
2 2 225 51 13 3 9 3
9 6 9 6 23
9 9 23
x x y y
x x y y
x x y y
x y
Yes, it is a circle with center 1 13 3
, and
radius 53
.
Section 2.2 Circles 183
Copyright © 2017 Pearson Education, Inc.
38. 2 24 4 4 4 7 0x y x y Complete the square on x and y separately.
2 2
2 21 14 4
1 14 4
2 21 12 2
2 2 91 12 2 4
4 4 7
4 4 –
7 4 4
4 4 – 9
–
x x y y
x x y y
x y
x y
Yes, it is a circle with center 1 12 2
, and
radius 32
.
39. The equations of the three circles are 2 2( 7) ( 4) 25x y , 2 2( 9) ( 4) 169x y , and 2 2( 3) ( 9) 100x y . From the graph of the
three circles, it appears that the epicenter is located at (3, 1).
Check algebraically:
2 2
2 2
2 2
( 7) ( 4) 25
(3 7) (1 4) 25
4 3 25 25 25
x y
2 2
2 2
2 2
( 9) ( 4) 169
(3 9) (1 4) 169
12 5 169 169 169
x y
2 2
2 2
2 2
( 3) ( 9) 100
(3 3) (1 9) 100
6 ( 8) 100 100 100
x y
(3, 1) satisfies all three equations, so the epicenter is at (3, 1).
40. The three equations are 2 2( 3) ( 1) 5x y , 2 2( 5) ( 4) 36x y , and 2 2( 1) ( 4) 40x y . From the graph of the
three circles, it appears that the epicenter is located at (5, 2).
Check algebraically:
2 2
2 2
2 2
( 3) ( 1) 5
(5 3) (2 1) 5
2 1 5 5 5
x y
2 2
2 2
2
( 5) ( 4) 36
(5 5) (2 4) 36
6 36 36 36
x y
2 2
2 2
2 2
( 1) ( 4) 40
(5 1) (2 4) 40
6 ( 2) 40 40 40
x y
(5, 2) satisfies all three equations, so the epicenter is at (5, 2).
41. From the graph of the three circles, it appears that the epicenter is located at (−2, −2).
Check algebraically:
2 2
2 2
2 2
( 2) ( 1) 25
( 2 2) ( 2 1) 25
( 4) ( 3) 2525 25
x y
2 2
2 2
2 2
( 2) ( 2) 16
( 2 2) ( 2 2) 16
0 ( 4) 1616 16
x y
2 2
2 2
2 2
( 1) ( 2) 9
( 2 1) ( 2 2) 9
( 3) 0 99 9
x y
(−2, −2) satisfies all three equations, so the epicenter is at (−2, −2).
184 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
42. From the graph of the three circles, it appears that the epicenter is located at (5, 0).
Check algebraically:
2 2
2 2
2 2
( 2) ( 4) 25
(5 2) (0 4) 25
3 ( 4) 2525 25
x y
2 2
2 2
2 2
( 1) ( 3) 25
(5 1) (0 3) 25
4 3 2525 25
x y
2 2
2 2
2 2
( 3) ( 6) 100
(5 3) (0 6) 100
8 6 100100 100
x y
(5, 0) satisfies all three equations, so the epicenter is at (5, 0).
43. The radius of this circle is the distance from the center C(3, 2) to the x-axis. This distance is 2, so r = 2.
2 2 2
2 2
( – 3) ( – 2) 2
( – 3) ( – 2) 4
x yx y
44. The radius is the distance from the center C(–4, 3) to the point P(5, 8).
2 2
2 2
[5 – (– 4)] (8 – 3)
9 5 106
r
The equation of the circle is 2 2 2
2 3
[ – (– 4)] ( – 3) ( 106)
( 4) ( – 3) 106
x yx y
45. Label the points P(x, y) and Q(1, 3).
If ( , ) 4d P Q , 2 21 3 4x y
2 21 3 16.x y
If x = y, then we can either substitute x for y or y for x. Substituting x for y we solve the following:
2 2
2 2
2
2
2
1 3 16
1 2 9 6 16
2 8 10 16
2 8 6 0
4 3 0
x xx x x x
x xx xx x
To solve this equation, we can use the quadratic formula with a = 1, b = –4, and c = –3.
24 4 4 1 3
2 1
4 16 12 4 28
2 2
4 2 72 7
2
x
Because x = y, the points are
2 7, 2 7 and 2 – 7, 2 7 .
46. Let P(–2, 3) be a point which is 8 units from Q(x, y). We have
2 2( , ) 2 3 8d P Q x y
2 22 3 64.x y
Because x + y = 0, x = –y. We can either substitute x for y or y for x. Substituting
x for y we solve the following:
22
2 2
2 2
2
2
2 3 64
2 3 64
4 4 9 6 64
2 10 13 64
2 10 51 0
x x
x xx x x x
x xx x
To solve this equation, use the quadratic formula with a = 2, b = 10, and c = –51.
210 10 4 2 51
2 2
10 100 408
410 4 12710 508
4 410 2 127 5 127
4 2
x
Because y x the points are
5 127 5 127,
2 2
and
–5 127 5 127, .
2 2
Section 2.2 Circles 185
Copyright © 2017 Pearson Education, Inc.
47. Let P(x, y) be a point whose distance from
A(1, 0) is 10 and whose distance from
B(5, 4) is 10 . d(P, A) = 10 , so
2 2(1 ) (0 ) 10x y 2 2(1 ) 10.x y ( , ) 10, sod P B
2 2(5 ) (4 ) 10x y 2 2(5 ) (4 ) 10.x y Thus,
2 2 2 2
2 2
2 2
(1 ) (5 ) (4 )
1 2
25 10 16 81 2 41 10 8
8 40 85
x y x yx x y
x x y yx x yy xy x
Substitute 5 – x for y in the equation 2 2(1 ) 10x y and solve for x. 2 2
2 2
2 2
2
(1 ) (5 ) 10
1 2 25 10 10
2 12 26 10 2 12 16 0
6 8 0 ( 2)( 4) 02 0 or 4 0
2 or 4
x xx x x x
x x x xx x x xx x
x x
To find the corresponding values of y use the equation y = 5 – x. If x = 2, then y = 5 – 2 = 3. If x = 4, then y = 5 – 4 = 1. The points satisfying the conditions are (2, 3) and (4, 1).
48. The circle of smallest radius that contains the points A(1, 4) and B(–3, 2) within or on its boundary will be the circle having points A and B as endpoints of a diameter. The center will be M, the midpoint:
1 3 4 2 2 6, ,
2 2 2 2
(−1, 3).
The radius will be the distance from M to either A or B:
2 2
2 2
( , ) [1 ( 1)] (4 3)
2 1 4 1 5
d M A
The equation of the circle is
22 2
2 2
1 3 5
1 3 5.
x y
x y
49. Label the points A(3, y) and B(–2, 9). If d(A, B) = 12, then
2 2
2 2
2 2 2
2
2
2 3 9 12
5 9 12
5 9 12
25 81 18 144
18 38 0
y
y
yy y
y y
Solve this equation by using the quadratic formula with a = 1, b = –18, and c = –38:
218 18 4 1 38
2 1
18 324 152 18 476
2 1 2
y
18 4 119 18 2 119
9 1192 2
The values of y are 9 119 and 9 119 .
50. Because the center is in the third quadrant, the
radius is 2 , and the circle is tangent to both
axes, the center must be at ( 2, 2). Using the center-radius of the equation of a circle, we have
2 2 2
2 2
2 2 2
2 2 2.
x y
x y
51. Let P(x, y) be the point on the circle whose distance from the origin is the shortest. Complete the square on x and y separately to write the equation in center-radius form:
2 2
2 2
2 2
16 14 88 0
16 64 14 4988 64 49
( 8) ( 7) 25
x x y yx x y y
x y
So, the center is (8, 7) and the radius is 5.
2 2( , ) 8 7 113d C O . Because the
length of the radius is 5, ( , ) 113 5d P O .
186 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
52. Using compasses, draw circles centered at Wickenburg, Kingman, Phoenix, and Las Vegas with scaled radii of 50, 75, 105, and 180 miles respectively. The four circles should intersect at the location of Nothing.
53. The midpoint M has coordinates
3 –9–1+5 4 6 , , (2, – 3).
2 2 2 2
54. Use points C(2, –3) and P(–1, 3).
22
2 2
( , ) –1 – 2 3 – –3
–3 6 9 36
45 3 5
d C P
The radius is 3 5.
55. Use points C(2, –3) and Q(5, –9).
22
22
( , ) 5 – 2 –9 – –3
3 –6 9 36
45 3 5
d C Q
The radius is 3 5.
56. Use the points P(–1, 3) and Q(5, –9).
Because 2 2( , ) 5 – –1 –9 – 3d P Q
226 –12 36 144 180
6 5 ,the radius is 1
( , ).2
d P Q Thus
16 5 3 5.
2r
57. The center-radius form for this circle is 2 2 2( – 2) ( 3) (3 5)x y 2 2( – 2) ( 3) 45.x y
58. Label the endpoints of the diameter P(3, –5) and Q(–7, 3). The midpoint M of the segment joining P and Q has coordinates
3 (–7) –5 3 4 –2, , (–2, –1).
2 2 2 2
The center is C(–2, –1). To find the radius, we can use points C(–2, –1) and P(3, –5)
2 2
22
( , ) 3 – –2 –5 – –1
5 –4 25 16 41
d C P
We could also use points C(–2, –1).and Q(–7, 3).
2 2
2 2
( , ) 7 – –2 3 – –1
5 4 25 16 41
d C Q
We could also use points P(3, –5) and Q(–7, 3) to find the length of the diameter. The length of the radius is one-half the length of the diameter.
22
2 2
( , ) 7 – 3 3 – –5
10 8 100 64
164 2 41
d P Q
1 1( , ) 2 41 41
2 2d P Q
The center-radius form of the equation of the circle is
2 2 2
2 2
[ – (–2)] [ – (–1)] ( 41)
( 2) ( 1) 41
x yx y
59. Label the endpoints of the diameter P(–1, 2) and Q(11, 7). The midpoint M of the segment joining P and Q has coordinates
1 11 2 7 9, 5, .
2 2 2
The center is 92
5, .C To find the radius, we
can use points 92
5,C and P(−1, 2).
22 92
22 5 169 132 4 2
( , ) 5 – –1 2
6
d C P
We could also use points 92
5,C and
Q(11, 7).
22 92
22 5 169 132 4 2
( , ) 5 11 – 7
6
d C Q
(continued on next page)
Section 2.3 Functions 187
Copyright © 2017 Pearson Education, Inc.
(continued)
Using the points P and Q to find the length of the diameter, we have
2 2
2 2
, 1 11 2 7
12 5
169 13
d P Q
1 1 13, 13
2 2 2d P Q
The center-radius form of the equation of the circle is
2 22 9 132 2
22 9 1692 4
5
5
x y
x y
60. Label the endpoints of the diameter P(5, 4) and Q(−3, −2). The midpoint M of the segment joining P and Q has coordinates
5 ( 3) 4 ( 2), 1, 1 .
2 2
The center is C(1, 1). To find the radius, we can use points C(1, 1) and P(5, 4).
2 2
2 2
( , ) 5 1 4 1
4 3 25 5
d C P
We could also use points C(1, 1) and Q(−3, −2).
2 2
2 2
( , ) 1 3 1 –2
4 3 25 5
d C Q
Using the points P and Q to find the length of the diameter, we have
2 2
2 2
, 5 3 4 2
8 6 100 10
d P Q
1 1( , ) 10 5
2 2d P Q
The center-radius form of the equation of the circle is
2 2 2
2 2
1 1 5
1 1 25
x yx y
61. Label the endpoints of the diameter P(1, 4) and Q(5, 1). The midpoint M of the segment joining P and Q has coordinates
52
1 5 4 1, 3, .
2 2
The center is 52
3, .C
The length of the diameter PQ is
2 2 2 21 5 4 1 4 3 25 5.
The length of the radius is 512 2
5 .
The center-radius form of the equation of the circle is
2 22 5 52 2
22 5 252 4
3
3
x y
x y
62. Label the endpoints of the diameter P(−3, 10) and Q(5, −5). The midpoint M of the segment joining P and Q has coordinates
52
3 5 10 ( 5), 1, .
2 2
The center is 52
1, .C
The length of the diameter PQ is
22 2 23 5 10 5 8 15
289 17.
The length of the radius is 1712 2
17 .
The center-radius form of the equation of the circle is
2 22 5 172 2
22 5 2892 4
1
1
x y
x y
Section 2.3 Functions
1. The domain of the relation
3,5 , 4,9 , 10,13 is 3, 4,10 .
2. The range of the relation in Exercise 1 is
5,9,13 .
3. The equation y = 4x – 6 defines a function with independent variable x and dependent variable y.
4. The function in Exercise 3 includes the ordered pair (6, 18).
5. For the function 4 2,f x x
2 4 2 2 8 2 10.f
6. For the function ,g x x 9 9 3.g
7. The function in Exercise 6, ,g x x has
domain 0, .
188 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
8. The function in Exercise 6, ,g x x has
range 0, .
For exercises 9 and 10, use this graph.
9. The largest open interval over which the function graphed here increases is , 3 .
10. The largest open interval over which the function graphed here decreases is 3, .
11. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
12. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
13. Two ordered pairs, namely (2, 4) and (2, 6), have the same x-value paired with different y-values, so the relation is not a function.
14. Two ordered pairs, namely (9, −2) and (9, 1), have the same x-value paired with different y-values, so the relation is not a function.
15. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
16. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
17. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
18. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
19. Two sets of ordered pairs, namely (1, 1) and (1, −1) as well as (2, 4) and (2, −4), have the same x-value paired with different y-values, so the relation is not a function. domain: {0, 1, 2}; range: {−4, −1, 0, 1, 4}
20. The relation is not a function because the x-value 3 corresponds to two y-values, 7 and 9. This correspondence can be shown as follows.
domain: {2, 3, 5}; range: {5, 7, 9, 11}
21. The relation is a function because for each different x-value there is exactly one y-value. domain: {2, 3, 5, 11, 17}; range: {1, 7, 20}
22. The relation is a function because for each different x-value there is exactly one y-value. domain: {1, 2, 3, 5}; range: {10, 15, 19, 27}
23. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
Domain: {0, −1, −2}; range: {0, 1, 2}
Section 2.3 Functions 189
Copyright © 2017 Pearson Education, Inc.
24. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.
Domain: {0, 1, 2}; range: {0, −1, −2}
25. The relation is a function because for each different year, there is exactly one number for visitors. domain: {2010, 2011, 2012, 2013} range: {64.9, 63.0, 65.1, 63.5}
26. The relation is a function because for each basketball season, there is only one number for attendance. domain: {2011, 2012, 2013, 2014} range: {11,159,999, 11,210,832, 11,339,285, 11,181,735}
27. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: ,
28. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: , 4
29. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: 3, ; range: ,
30. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: [−4, 4]; range: [−3, 3]
31. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: ,
32. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: [−2, 2]; range: [0, 4]
33. 2y x represents a function because y is always found by squaring x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the square of any real number is not negative, the range would be zero or greater.
domain: , ; range: 0,
34. 3y x represents a function because y is always found by cubing x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the cube of any real number could be negative, positive, or zero, the range would be any real number.
domain: , ; range: ,
35. The ordered pairs (1, 1) and (1, −1) both
satisfy 6.x y This equation does not represent a function. Because x is equal to the sixth power of y, the values of x are nonnegative. Any real number can be raised to the sixth power, so the range of the relation is all real numbers.
domain: 0, range: ,
190 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
36. The ordered pairs (1, 1) and (1, −1) both
satisfy 4.x y This equation does not represent a function. Because x is equal to the fourth power of y, the values of x are nonnegative. Any real number can be raised to the fourth power, so the range of the relation is all real numbers.
domain: 0, range: ,
37. 2 5y x represents a function because y is found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers.
domain: , ; range: ,
38. 6 4y x represents a function because y is found by multiplying x by −6 and adding 4. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is −6 times x, plus 4, y also may be any real number, and so the range is also all real numbers.
domain: , ; range: ,
39. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x + y < 3 corresponds to many values of y. The ordered pairs (1, –2), (1, 1), (1, 0), (1, −1), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.
domain: , ; range: ,
40. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x − y < 4 corresponds to many values of y. The ordered pairs (1, −1), (1, 0), (1, 1), (1, 2), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.
domain: , ; range: ,
Section 2.3 Functions 191
Copyright © 2017 Pearson Education, Inc.
41. For any choice of x in the domain of ,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x Because the radical is nonnegative, the range is also zero or greater.
domain: 0, ; range: 0,
42. For any choice of x in the domain of
,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x The outcome of the radical is nonnegative, when you change the sign (by multiplying by −1), the range becomes nonpositive. Thus the range is zero or less.
domain: 0, ; range: , 0
43. Because 2xy can be rewritten as 2 ,xy
we can see that y can be found by dividing x into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: , 0 0, ;
range: , 0 0,
44. Because xy = −6 can be rewritten as 6 ,xy
we can see that y can be found by dividing x into −6. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: , 0 0, ;
range: , 0 0,
45. For any choice of x in the domain of
4 1y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have
4 1 0 4 1x x 14
.x Because the
radical is nonnegative, the range is also zero or greater.
domain: 1
4, ; range: 0,
192 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
46. For any choice of x in the domain of
7 2y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have
7 2 0 2 7x x 7 72 2
or .x x
Because the radical is nonnegative, the range is also zero or greater.
domain: 72
, ; range: 0,
47. Given any value in the domain of 23
,xy we
find y by subtracting 3, then dividing into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 3. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: ,3 3, ;
range: , 0 0,
48. Given any value in the domain of 75
,xy we
find y by subtracting 5, then dividing into −7. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 5. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.
domain: ,5 5, ;
range: , 0 0,
49. B. The notation f(3) means the value of the dependent variable when the independent variable is 3.
50. Answers will vary. An example is: The cost of gasoline depends on the number of gallons used; so cost is a function of number of gallons.
51.
3 4
0 3 0 4 0 4 4
f x xf
52.
3 4
3 3 3 4 9 4 13
f x xf
53.
2
2
4 1
2 2 4 2 1
4 8 1 11
g x x xg
54.
2
2
4 1
10 10 4 10 1100 40 1 59
g x x xg
55. 1 1
3 3
3 4
3 4 1 4 3
f x xf
56. 7 7
3 3
3 4
3 4 7 4 11
f x xf
57.
2
21 1 12 2 2
1 114 4
4 1
4 1
2 1
g x x x
g
58.
2
21 1 14 4 4
1 116 16
4 1
4 1
1 1
g x x x
g
59.
3 4
3 4
f x xf p p
Section 2.3 Functions 193
Copyright © 2017 Pearson Education, Inc.
60.
2
2
4 1
4 1
g x x xg k k k
61.
3 4
3 4 3 4
f x xf x x x
62.
2
2
2
4 1
4 1
4 1
g x x xg x x x
x x
63.
3 4
2 3 2 43 6 4 3 2
f x xf x x
x x
64.
3 4
4 3 4 43 12 4 3 8
f x xf a a
a a
65.
3 4
2 3 3 2 3 46 9 4 6 13
f x xf m m
m m
66.
3 4
3 2 3 3 2 49 6 4 9 10
f x xf t t
t t
67. (a) 2 2f (b) 1 3f
68. (a) 2 5f (b) 1 11f
69. (a) 2 15f (b) 1 10f
70. (a) 2 1f (b) 1 7f
71. (a) 2 3f (b) 1 3f
72. (a) 2 3f (b) 1 2f
73. (a) 2 0f (b) 0 4f
(c) 1 2f (d) 4 4f
74. (a) 2 5f (b) 0 0f
(c) 1 2f (d) 4 4f
75. (a) 2 3f (b) 0 2f
(c) 1 0f (d) 4 2f
76. (a) 2 3f (b) 0 3f
(c) 1 3f (d) 4 3f
77. (a)
1 13 3
3 123 12
12
34 4
x yy x
xy
y x f x x
(b) 13
3 3 4 1 4 3f
78. (a)
1 14 4
4 88 4
8 48
42 2
x yx y
x yx y
y x f x x
(b) 3 3 8 514 4 4 4 4
3 3 2 2f
79. (a)
2
2
2
2 3
2 3
2 3
y x xy x x
f x x x
(b) 23 2 3 3 32 9 3 3 18
f
80. (a)
2
2
2
3 2
3 2
3 2
y x xy x x
f x x x
(b) 23 3 3 3 23 9 3 2 32
f
81. (a)
8 84 43 3 3 3
4 3 84 3 8
4 8 34 8
3
x yx y
x yx y
y x f x x
(b) 8 84 12 43 3 3 3 3
3 3f
82. (a)
9 92 25 5 5 5
2 5 95 2 9
2 9
5
x yy x
xy
y x f x x
(b) 9 6 9 1525 5 5 5 5
3 3 3f
83. 3 4f
194 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
84. Because 20.2 0.2 3 0.2 1f
= 0.04 + 0.6 + 1 = 1.64, the height of the rectangle is 1.64 units. The base measures 0.3 − 0.2 = 0.1 unit. Because the area of a rectangle is base times height, the area of this rectangle is 0.1(1.64) = 0.164 square unit.
85. 3f is the y-component of the coordinate,
which is −4.
86. 2f is the y-component of the coordinate,
which is −3.
87. (a) 2, 0 (b) , 2
(c) 0,
88. (a) 3, 1 (b) 1,
(c) , 3
89. (a) , 2 ; 2,
(b) 2, 2 (c) none
90. (a) 3,3 (b) , 3 ; 3,
(c) none
91. (a) 1, 0 ; 1,
(b) , 1 ; 0, 1
(c) none
92. (a) , 2 ; 0, 2
(b) 2, 0 ; 2,
(c) none
93. (a) Yes, it is the graph of a function.
(b) [0, 24]
(c) When t = 8, y = 1200 from the graph. At 8 A.M., approximately 1200 megawatts is being used.
(d) The most electricity was used at 17 hr or 5 P.M. The least electricity was used at 4 A.M.
(e) 12 1900f
At 12 noon, electricity use is about 1900 megawatts.
(f) increasing from 4 A.M. to 5 P.M.; decreasing from midnight to 4 A.M. and from 5 P.M. to midnight
94. (a) At t = 2, y = 240 from the graph. Therefore, at 2 seconds, the ball is 240 feet high.
(b) At y = 192, x = 1 and x = 5 from the graph. Therefore, the height will be 192 feet at 1 second and at 5 seconds.
(c) The ball is going up from 0 to 3 seconds and down from 3 to 7 seconds.
(d) The coordinate of the highest point is (3, 256). Therefore, it reaches a maximum height of 256 feet at 3 seconds.
(e) At x = 7, y = 0. Therefore, at 7 seconds, the ball hits the ground.
95. (a) At t = 12 and t = 20, y = 55 from the graph. Therefore, after about 12 noon until about 8 P.M. the temperature was over 55º.
(b) At t = 6 and t = 22, y = 40 from the graph. Therefore, until about 6 A.M. and after 10 P.M. the temperature was below 40º.
(c) The temperature at noon in Bratenahl, Ohio was 55º. Because the temperature in Greenville is 7º higher, we are looking for the time at which Bratenahl, Ohio was at or above 48º. This occurred at approximately 10 A.M and 8:30 P.M.
(d) The temperature is just below 40° from midnight to 6 A.M., when it begins to rise until it reaches a maximum of just below 65° at 4 P.M. It then begins to fall util it reaches just under 40° again at midnight.
96. (a) At t = 8, y = 24 from the graph. Therefore, there are 24 units of the drug in the bloodstream at 8 hours.
(b) The level increases between 0 and 2 hours after the drug is taken and decreases between 2 and 12 hours after the drug is taken.
(c) The coordinates of the highest point are (2, 64). Therefore, at 2 hours, the level of the drug in the bloodstream reaches its greatest value of 64 units.
(d) After the peak, y = 16 at t = 10. 10 hours – 2 hours = 8 hours after the peak. 8 additional hours are required for the level to drop to 16 units.
Section 2.4 Linear Functions 195
Copyright © 2017 Pearson Education, Inc.
(e) When the drug is administered, the level is 0 units. The level begins to rise quickly for 2 hours until it reaches a maximum of 64 units. The level then begins to decrease gradually until it reaches a level of 12 units, 12 hours after it was administered.
Section 2.4 Linear Functions
1. B; ( )f x = 3x + 6 is a linear function with y-intercept (0, 6).
2. H; x = 9 is a vertical line.
3. C; ( )f x = −8 is a constant function.
4. G; 2x – y = –4 or y = 2x + 4 is a linear equation with x-intercept (–2, 0) and y-intercept (0, 4).
5. A; ( )f x = 5x is a linear function whose graph passes through the origin, (0, 0). f(0) = 5(0) = 0.
6. D; 2( )f x x is a function that is not linear.
7. –3m matches graph C because the line falls rapidly as x increases.
8. m = 0 matches graph A because horizontal lines have slopes of 0.
9. m = 3 matches graph D because the line rises rapidly as x increases.
10. m is undefined for graph B because vertical lines have undefined slopes.
11. ( ) – 4f x x Use the intercepts.
(0) 0 – 4 – 4 :f y-intercept
0 – 4 4 :x x x-intercept Graph the line through (0, –4) and (4, 0).
The domain and range are both , .
12. ( ) – 4f x x Use the intercepts.
(0) –0 4 4 :f y-intercept
0 – 4 4 :x x x-intercept Graph the line through (0, 4) and (4, 0).
The domain and range are both , .
13. 12
( ) – 6f x x
Use the intercepts.
12
(0) 0 – 6 – 6 :f y-intercept
1 12 2
0 – 6 6 12 :x x x x-intercept
Graph the line through (0, –6) and (12, 0).
The domain and range are both , .
14. 23
( ) 2f x x
Use the intercepts.
23
(0) 0 2 2 :f y-intercept
2 23 3
0 2 –2 3 : -interceptx x x x
Graph the line through (0, 2) and (–3, 0).
The domain and range are both , .
196 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
15. ( ) 3f x x The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 1, y = 3(1) = 3 . Another point is (1, 3). Graph the line through (0, 0) and (1, 3).
The domain and range are both , .
16. –2f x x
The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 3, y = –2(3) = –6, so another point is (3, –6). Graph the line through (0, 0) and (3, –6).
The domain and range are both , .
17. ( ) – 4f x is a constant function.
The graph of ( ) 4f x is a horizontal line with a y-intercept of –4.
domain: , ; range: {–4}
18. ( ) 3f x is a constant function whose graph is a horizontal line with y-intercept of 3.
domain: , ; range: {3}
19. ( ) 0f x is a constant function whose graph is the x-axis.
domain: , ; range: {0}
20. 9f x x
The domain and range are both , .
21. 4 3 12x y Use the intercepts.
4 0 3 12 3 124 : -intercept
y yy y
4 3 0 12 4 123 : -intercept
x xx x
Graph the line through (0, 4) and (−3, 0).
The domain and range are both , .
22. 2 5 10;x y Use the intercepts.
2 0 5 10 5 102 : -intercept
y yy y
2 5 0 10 2 105 : -intercept
x xx x
Graph the line through (0, 2) and (5, 0):
The domain and range are both , .
Section 2.4 Linear Functions 197
Copyright © 2017 Pearson Education, Inc.
23. 3 4 0y x Use the intercepts.
3 4 0 0 3 0 0 : -intercepty y y y
3 0 4 0 4 0 0 : -interceptx x x x
The graph has just one intercept. Choose an additional value, say 3, for x.
3 4 3 0 3 12 03 12 4
y yy y
Graph the line through (0, 0) and (3, 4):
The domain and range are both , .
24. 3 2 0x y Use the intercepts. 3 0 2 0 2 0 0 : -intercepty y y y
3 2 0 0 3 0 0 :x x x x-intercept
The graph has just one intercept. Choose an additional value, say 2, for x.
3 2 2 0 6 2 02 6 3
y yy y
Graph the line through (0, 0) and (2, −3):
The domain and range are both , .
25. x = 3 is a vertical line, intersecting the x-axis at (3, 0).
domain: {3}; range: ,
26. x = –4 is a vertical line intersecting the x-axis at (–4, 0).
domain: {−4}; range: ,
27. 2x + 4 = 0 2 4 2x x is a vertical line intersecting the x-axis at (–2, 0).
domain: {−2}; range: ,
28. 3 6 0 –3 6 2x x x is a vertical
line intersecting the x-axis at (2, 0).
domain: {2}; range: ,
29. 5 0 5x x is a vertical line intersecting the x-axis at (5, 0).
domain: {5}; range: ,
198 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
30. 3 0 3x x is a vertical line
intersecting the x-axis at 3,0 .
domain: {−3}; range: ,
31. y = 5 is a horizontal line with y-intercept 5. Choice A resembles this.
32. y = –5 is a horizontal line with y-intercept –5. Choice C resembles this.
33. x = 5 is a vertical line with x-intercept 5. Choice D resembles this.
34. x = –5 is a vertical line with x-intercept –5. Choice B resembles this.
35.
36.
37.
38.
39. The rise is 2.5 feet while the run is 10 feet so
the slope is 2.5 110 4
0.25 25% . So A =
0.25, 2.510
C , D = 25%, and 14
E are all
expressions of the slope.
40. The pitch or slope is 14
. If the rise is 4 feet
then 1 rise 4
4 run x or x = 16 feet. So 16 feet in
the horizontal direction corresponds to a rise of 4 feet.
41. Through (2, –1) and (–3, –3) Let 1 1 22, –1, –3, and x y x 2 –3.y
Then rise –3 – (–1) –2y and
run –3 – 2 –5.x
The slope is rise –2 2
.run –5 5
ymx
42. Through (−3, 4) and (2, −8) Let 1 1 2 23, 4, 2, and –8.x y x y
Then rise –8 – 4 –12y and
run 2 – ( 3) 5.x
The slope is rise –12 12
.run 5 5
ymx
43. Through (5, 8) and (3, 12) Let 1 1 2 25, 8, 3, and 12.x y x y
Then rise 12 8 4y and
run 3 5 2.x
The slope is rise 4
2.run 2
ymx
44. Through (5, –3) and (1, –7)
1 1 2 2Let 5, –3, 1, and –7.x y x y
Then rise –7 – (–3) – 4y and
run 1 – 5 – 4.x
The slope is 4
1.4
ymx
45. Through (5, 9) and (–2, 9)
2 1
2 1
9 – 9 00
–2 5 –7
y yymx x x
46. Through (–2, 4) and (6, 4)
2 1
2 1
4 4 00
6 (–2) 8
y yymx x x
47. Horizontal, through (5, 1) The slope of every horizontal line is zero, so m = 0.
48. Horizontal, through (3 , 5) The slope of every horizontal line is zero, so m = 0.
49. Vertical, through (4, –7) The slope of every vertical line is undefined; m is undefined.
Section 2.4 Linear Functions 199
Copyright © 2017 Pearson Education, Inc.
50. Vertical, through (–8, 5) The slope of every vertical line is undefined; m is undefined.
51. (a) 3 5y x Find two ordered pairs that are solutions to the equation. If x = 0, then 3 0 5 5.y y
If x = −1, then 3 1 5 3 5 2.y y y
Thus two ordered pairs are (0, 5) and (−1, 2)
2 1
2 1
rise 2 5 33.
run 1 0 1
y ymx x
(b)
52. 2 4y x Find two ordered pairs that are solutions to the equation. If 0,x then 2 0 4y
4.y If 1,x then 2 1 4y
2 4 2.y y Thus two ordered pairs
are 0, 4 and 1, 2 .
2 1
2 1
2 4rise 22.
run 1 0 1
y ymx x
(b)
53. 2 3y x Find two ordered pairs that are solutions to the equation. If 0,x then 2 0 0.y y
If 3,y then 2 3 3 6 3x x
2.x Thus two ordered pairs are 0,0 and
2, 3 .
2 1
2 1
rise 3 0 3.
run 2 0 2
y ymx x
(b)
54. 4 5y x Find two ordered pairs that are solutions to the equation. If 0,x then 4 0 0.y y
If 4,x then 4 5 4 4 20y y
5.y Thus two ordered pairs are 0,0 and
4, 5 .
2 1
2 1
rise 5 0 5.
run 4 0 4
y ymx x
(b)
55. 5 2 10x y Find two ordered pairs that are solutions to the equation. If 0,x then 5 0 2 10y
5.y If 0,y then 5 2 0 10x
5 10x 2.x
Thus two ordered pairs are 0, 5 and 2,0 .
2 1
2 1
0 5rise 5.
run 2 0 2
y ymx x
(b)
200 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
56. 4 3 12x y Find two ordered pairs that are solutions to the equation. If 0,x then 4 0 3 12y
3 12y 4.y If 0,y then
4 3 0 12 4 12x x 3.x Thus two
ordered pairs are 0, 4 and 3,0 .
2 1
2 1
rise 0 4 4.
run 3 0 3
y ymx x
(b)
57. Through (–1, 3), 32
m
First locate the point (–1, 3). Because the
slope is 32
, a change of 2 units horizontally (2
units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (1, 6), which can be used to complete the graph.
58. Through (–2, 8), 25
m . Because the slope is
25
, a change of 5 units horizontally (to the
right) produces a change of 2 units vertically (2 units up). This gives a second point (3, 10), which can be used to complete the graph. Alternatively, a change of 5 units to the left produces a change of 2 units down. This gives the point (−7, 6).
59. Through (3, –4), 13
–m . First locate the point
(3, –4). Because the slope is 13
– , a change of
3 units horizontally (3 units to the right) produces a change of –1 unit vertically (1 unit down). This gives a second point, (6, –5), which can be used to complete the graph.
60. Through (–2, –3), 34
–m . Because the slope
is 3 –34 4
– , a change of 4 units horizontally
(4 units to the right) produces a change of –3 units vertically (3 units down). This gives a second point (2, –6), which can be used to complete the graph.
61. Through 12
, 4 , m = 0.
The graph is the horizontal line through
12
, 4 .
Exercise 61 Exercise 62
62. Through 32
, 2 , m = 0.
The graph is the horizontal line through
32
, 2 .
Section 2.4 Linear Functions 201
Copyright © 2017 Pearson Education, Inc.
63. Through 52
, 3 , undefined slope. The slope
is undefined, so the line is vertical,
intersecting the x-axis at 52
, 0 .
64. Through 94
, 2 , undefined slope. The slope is
undefined, so the line is vertical, intersecting
the x-axis at 94
, 0 .
65. The average rate of change is f b f a
mb a
20 4 16$4
0 4 4
(thousand) per year. The
value of the machine is decreasing $4000 each year during these years.
66. The average rate of change is f b f a
mb a
200 0 200$50
4 0 4
per month. The amount
saved is increasing $50 each month during these months.
67. The graph is a horizontal line, so the average rate of change (slope) is 0. The percent of pay raise is not changing—it is 3% each year.
68. The graph is a horizontal line, so the average rate of change (slope) is 0. That means that the number of named hurricanes remained the same, 10, for the four consecutive years shown.
69. 2562 5085 2523
2012 1980 3278.8 thousand per year
f b f am
b a
The number of high school dropouts decreased by an average of 78.8 thousand per year from 1980 to 2012.
70. 1709 5302
2013 20063593
$513.297
f b f am
b a
Sales of plasma flat-panel TVs decreased by an average of $513.29 million per year from 2006 to 2013.
71. (a) The slope of –0.0167 indicates that the average rate of change of the winning time for the 5000 m run is 0.0167 min less. It is negative because the times are generally decreasing as time progresses.
(b) The Olympics were not held during World Wars I (1914−1919) and II (1939–1945).
(c) 0.0167 2000 46.45 13.05 miny
The model predicts a winning time of 13.05 minutes. The times differ by 13.35 − 13.05 = 0.30 min.
72. (a) From the equation, the slope is 200.02. This means that the number of radio stations increased by an average of 200.02 per year.
(b) The year 2018 is represented by x = 68. 200.02 68 2727.7 16,329.06y
According to the model, there will be about 16,329 radio stations in 2018.
73. 2013 2008 335,652 270,334
2013 2008 2013 200865,318
13,063.65
f f
The average annual rate of change from 2008 through 2013 is about 13,064 thousand.
74. 2014 2006 3.74 4.53
2014 2006 2014 20060.79
0.0998
f f
The average annual rate of change from 2006 through 2014 is about –0.099.
202 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
75. (a) 56.3 138
2013 200381.7
8.1710
f b f am
b a
The average rate of change was −8.17 thousand mobile homes per year.
(b) The negative slope means that the number of mobile homes decreased by an average of 8.17 thousand each year from 2003 to 2013.
76. 2013 1991 26.6 61.8
2013 1991 2013 199135.2
1.622
f f
There was an average decrease of 1.6 births per thousand per year from 1991 through 2013.
77. (a) 10 500C x x
(b) 35R x x
(c) 35 10 500
35 10 500 25 500
P x R x C xx xx x x
(d) 10 500 35
500 2520
C x R xx x
xx
20 units; do not produce
78. (a) 150 2700C x x
(b) 280R x x
(c) 280 150 2700
280 150 2700130 2700
P x R x C xx xx xx
(d) 150 2700 280
2700 13020.77 or 21 units
C x R xx x
xx
21 units; produce
79. (a) 400 1650C x x
(b) 305R x x
(c) 305 400 1650
305 400 165095 1650
P x R x C xx xx xx
(d) 400 1650 305
95 1650 095 1650
17.37 units
C x R xx xx
xx
This result indicates a negative “break-even point,” but the number of units produced must be a positive number. A calculator graph of the lines
1 400 1650y C x x and
2 305y R x x in the window
[0, 70] × [0, 20000] or solving the inequality 305 400 1650x x will show
that R x C x for all positive values
of x (in fact whenever x is greater than −17.4). Do not produce the product because it is impossible to make a profit.
80. (a) 11 180C x x
(b) 20R x x
(c) 20 11 180
20 11 180 9 180
P x R x C xx xx x x
(d) 11 180 20
180 920
C x R xx x
xx
20 units; produce
81. 200 1000 2401000 40 25C x R x x x
x x
The break-even point is 25 units. (25) 200 25 1000 $6000C which is the
same as (25) 240 25 $6000R
C x
R x
Chapter 2 Quiz (Sections 2.1–2.4) 203
Copyright © 2017 Pearson Education, Inc.
82. 220 1000 2401000 20 50C x R x x x
x x
The break-even point is 50 units instead of 25 units. The manager is not better off because twice as many units must be sold before beginning to show a profit.
83. The first two points are A(0, –6) and B(1, –3). 3 – (–6) 3
31 – 0 1
m
84. The second and third points are B(1, –3) and C(2, 0).
0 – (–3) 33
2 – 1 1m
85. If we use any two points on a line to find its slope, we find that the slope is the same in all cases.
86. The first two points are A(0, –6) and B(1, –3). 2 2
2 2
( , ) (1 – 0) [–3 – (– 6)]
1 3 1 9 10
d A B
87. The second and fourth points are B(1, –3) and D(3, 3).
2 2
2 2
( , ) (3 – 1) [3 – (–3)]
2 6 4 36
40 2 10
d B D
88. The first and fourth points are A(0, –6) and D(3, 3).
2 2
2 2
( , ) (3 – 0) [3 – (– 6)]
3 9 9 81
90 3 10
d A D
89. 10 2 10 3 10; The sum is 3 10, which is equal to the answer in Exercise 88.
90. If points A, B, and C lie on a line in that order, then the distance between A and B added to the distance between B and C is equal to the distance between A and C.
91. The midpoint of the segment joining A(0, –6) and G(6, 12) has coordinates
0 6 6 122 2
, 6 62 2
, 3,3 . The midpoint is
M(3, 3), which is the same as the middle entry in the table.
92. The midpoint of the segment joining E(4, 6) and F(5, 9) has coordinates
4 5 6 9 9 152 2 2 2
, , = (4.5, 7.5). If the
x-value 4.5 were in the table, the corresponding y-value would be 7.5.
Chapter 2 Quiz (Sections 2.1−2.4)
1.
2 22 1 2 1
2 2
2 2
( , )
8 ( 4) 3 2
( 4) ( 5) 16 25 41
d A B x x y y
2. To find an estimate for 2006, find the midpoint of (2004, 6.55) and (2008, 6.97:
2004 2008 6.55 6.97,
2 22006, 6.76
M
The estimated enrollment for 2006 was 6.76 million. To find an estimate for 2010, find the midpoint of (2008, 6.97) and (2012, 7.50):
2008 2012 6.97 7.50,
2 22010, 7.235
M
The estimated enrollment for 2006 was about 7.24 million.
3.
4.
5. 2 2 4 8 3 0x y x y Complete the square on x and y separately.
2 2
2 2
( 4 4) ( 8 16) 3 4 16
( 2) ( 4) 17
x x y yx y
The radius is 17 and the midpoint of the circle is (2, −4).
6. From the graph, f(−1) is 2.
7. Domain: ( , ); range: [0, )
204 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
8. (a) The largest open interval over which f is decreasing is ( , 3) .
(b) The largest open interval over which f is increasing is 3, .
(c) There is no interval over which the function is constant.
9. (a) 11 5 6 3
5 1 4 2m
(b) 4 4 0
01 ( 7) 6
m
(c) 4 12 16
6 6 0m
the slope is
undefined.
10. The points to use are (2009, 10,602) and (2013, 15,884). The average rate of change is
15,884 10,602 5282
1320.52013 2009 4
The average rate of change was 1320.5 thousand cars per year. This means that the number of new motor vehicles sold in the United States increased by an average of 1320.5 thousand per year from 2009 to 2013.
Section 2.5 Equations of Lines and Linear Models
1. The graph of the line 3 4 8y x has
slope 4 and passes through the point (8, 3).
2. The graph of the line 2 7y x has slope –2 and y-intercept (0, 7).
3. The vertical line through the point (–4, 8) has equation x = –4.
4. The horizontal line through the point (–4, 8) has equation y = 8.
For exercises 5 and 6, 67
6 7 0 7 6x y y x y x
5. Any line parallel to the graph of 6 7 0x y
must have slope 67.
6. Any line perpendicular to the graph of
6 7 0x y must have slope 76.
7. 14
2y x is graphed in D.
The slope is 14
and the y-intercept is (0, 2).
8. 4x + 3y = 12 or 3y = –4x + 12 43
or – 4y x
is graphed in B. The slope is 43
– and the
y-intercept is (0, 4).
9. 32
– 1 1y x is graphed in C. The slope
is 32
and a point on the graph is (1, –1).
10. y = 4 is graphed in A. y = 4 is a horizontal line with y-intercept (0, 4).
11. Through (1, 3), m = –2. Write the equation in point-slope form.
1 1– – 3 –2 –1y y m x x y x
Then, change to standard form. 3 –2 2 2 5y x x y
12. Through (2, 4), m = –1 Write the equation in point-slope form.
1 1– – 4 –1 – 2y y m x x y x
Then, change to standard form. – 4 – 2 6y x x y
13. Through (–5, 4), 32
m
Write the equation in point-slope form.
32
– 4 5y x
Change to standard form. 2 – 4 –3 52 – 8 –3 – 15
3 2 –7
y xy x
x y
14. Through 34
(– 4, 3), m
Write the equation in point-slope form.
34
– 3 4y x
Change to standard form. 4 3 3 4
4 12 3 123 4 24 or 3 4 24
y xy xx y x y
15. Through (–8, 4), undefined slope Because undefined slope indicates a vertical line, the equation will have the form x = a. The equation of the line is x = –8.
16. Through (5, 1), undefined slope This is a vertical line through (5, 1), so the equation is x = 5.
17. Through (5, −8), m = 0 This is a horizontal line through (5, −8), so the equation is y = −8.
18. Through (−3, 12), m = 0 This is a horizontal line through (−3, 12), so the equation is y = 12.
Section 2.5 Equations of Lines and Linear Models 205
Copyright © 2017 Pearson Education, Inc.
19. Through (–1, 3) and (3, 4) First find m.
4 – 3 1
3 – (–1) 4m
Use either point and the point-slope form.
14
– 4 – 34 16 3
4 134 13
y xy xx yx y
20. Through (2, 3) and (−1, 2) First find m.
2 3 1 1
1 2 3 3m
Use either point and the point-slope form.
13
3 – 23 – 9 2
3 73 7
y xy x
x yx y
21. x-intercept (3, 0), y-intercept (0, –2) The line passes through (3, 0) and (0, –2). Use these points to find m.
–2 – 0 2
0 – 3 3m
Using slope-intercept form we have 23
– 2.y x
22. x-intercept (–4, 0), y-intercept (0, 3) The line passes through the points (–4, 0) and (0, 3). Use these points to find m.
3 – 0 3
0 – (–4) 4m
Using slope-intercept form we have 34
3.y x
23. Vertical, through (–6, 4) The equation of a vertical line has an equation of the form x = a. Because the line passes through (–6, 4), the equation is x = –6. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)
24. Vertical, through (2, 7) The equation of a vertical line has an equation of the form x = a. Because the line passes through (2, 7), the equation is x = 2. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)
25. Horizontal, through (−7, 4) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−7, 4), the equation is y = 4.
26. Horizontal, through (−8, −2) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−8, −2), the equation is y = −2.
27. m = 5, b = 15 Using slope-intercept form, we have
5 15.y x
28. m = −2, b = 12 Using slope-intercept form, we have
2 12.y x
29. Through (−2, 5), slope = −4
5 4 2
5 4 25 4 8
4 3
y xy xy x
y x
30. Through (4, −7), slope = −2 7 2 4
7 2 82 1
y xy x
y x
31. slope 0, y-intercept 32
0,
These represent 32
0 and .m b
Using slope-intercept form, we have
3 32 2
0 .y x y
32. slope 0, y-intercept 54
0,
These represent 54
0 and .m b
Using slope-intercept form, we have
5 54 4
0 .y x y
33. The line x + 2 = 0 has x-intercept (–2, 0). It does not have a y-intercept. The slope of his line is undefined. The line 4y = 2 has
y-intercept 12
0, . It does not have an
x-intercept. The slope of this line is 0.
34. (a) The graph of y = 3x + 2 has a positive slope and a positive y-intercept. These conditions match graph D.
206 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b) The graph of y = –3x + 2 has a negative slope and a positive y-intercept. These conditions match graph B.
(c) The graph of y = 3x – 2 has a positive slope and a negative y-intercept. These conditions match graph A.
(d) The graph of y = –3x – 2 has a negative slope and a negative y-intercept. These conditions match graph C.
35. y = 3x – 1 This equation is in the slope-intercept form, y = mx + b. slope: 3; y-intercept: (0, –1)
36. y = –2x + 7 slope: –2; y-intercept: (0, 7)
37. 4x – y = 7 Solve for y to write the equation in slope-intercept form. – – 4 7 4 – 7y x y x slope: 4; y-intercept: (0, –7)
38. 2x + 3y = 16 Solve the equation for y to write the equation in slope-intercept form.
1623 3
3 –2 16 –y x y x
slope: 23
– ; y-intercept: 163
0,
39. 3 34 4
4 –3 – or 0y x y x y x
slope: 34
; y-intercept (0, 0)
40. 1 12 2
2 or 0y x y x y x
slope is 12
; y-intercept: (0, 0)
41. 2 4x y Solve the equation for y to write the equation in slope-intercept form.
12
2 – 4 – 2y x y x
slope: 12
– ; y-intercept: (0, −2)
Section 2.5 Equations of Lines and Linear Models 207
Copyright © 2017 Pearson Education, Inc.
42. 3 9x y Solve the equation for y to write the equation in slope-intercept form.
13
3 – 9 – 3y x y x
slope: 13
– ; y-intercept: (0,−3)
43. 32
1 0y x
Solve the equation for y to write the equation in slope-intercept form.
3 32 2
1 0 1y x y x
slope: 32
; y-intercept: (0, 1)
44. (a) Use the first two points in the table, A(–2, –11) and B(–1, –8).
–8 – (–11) 33
–1 – (–2) 1m
(b) When x = 0, y = −5. The y-intercept is (0, –5).
(c) Substitute 3 for m and –5 for b in the slope-intercept form.
3 – 5y mx b y x
45. (a) The line falls 2 units each time the x value increases by 1 unit. Therefore the slope is −2. The graph intersects the y-axis at the point (0, 1) and intersects the
x-axis at 12
, 0 , so the y-intercept is
(0, 1) and the x-intercept is 12
, 0 .
(b) An equation defining f is y = −2x + 1.
46. (a) The line rises 2 units each time the x value increases by 1 unit. Therefore the slope is 2. The graph intersects the y-axis at the point (0, −1) and intersects the
x-axis at 12
, 0 , so the y-intercept is
(0, −1) and the x-intercept is 12
, 0 .
(b) An equation defining f is y = 2x − 1.
47. (a) The line falls 1 unit each time the x value increases by 3 units. Therefore the slope
is 13
. The graph intersects the y-axis at
the point (0, 2), so the y-intercept is (0, 2). The graph passes through (3, 1) and will fall 1 unit when the x value increases by 3, so the x-intercept is (6, 0).
(b) An equation defining f is 13
2.y x
48. (a) The line rises 3 units each time the x value increases by 4 units. Therefore the
slope is 34
. The graph intersects the
y-axis at the point (0, −3) and intersects the x-axis at (4, 0), so the y-intercept is (0, −3) and the x-intercept is 4.
(b) An equation defining f is 34
3.y x
49. (a) The line falls 200 units each time the x value increases by 1 unit. Therefore the slope is −200. The graph intersects the y-axis at the point (0, 300) and intersects
the x-axis at 32
, 0 , so the y-intercept is
(0, 300) and the x-intercept is 32
, 0 .
(b) An equation defining f is y = −200x + 300.
50. (a) The line rises 100 units each time the x value increases by 5 units. Therefore the slope is 20. The graph intersects the y-axis at the point (0, −50) and intersects
the x-axis at 52
, 0 , so the y-intercept is
(0, −50) and the x-intercept is 52
, 0 .
(b) An equation defining f is y = 20x − 50.
208 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
51. (a) through (–1, 4), parallel to x + 3y = 5 Find the slope of the line x + 3y = 5 by writing this equation in slope-intercept form.
513 3
3 5 3 – 5–
x y y xy x
The slope is 13
.
Because the lines are parallel, 13
is also
the slope of the line whose equation is to
be found. Substitute 13
m , 1 –1,x
and 1 4y into the point-slope form.
1 11313
– –
4 – –1
– 4 13 –12 – –1 3 11
y y m x xy xy x
y x x y
(b) Solve for y. 1 113 3
3 – 11 –y x y x
52. (a) through (3, –2), parallel to 2x – y = 5 Find the slope of the line 2x – y = 5 by writing this equation in slope-intercept form. 2 – 5 – –2 5
2 – 5x y y x
y x
The slope is 2. Because the lines are parallel, the slope of the line whose equation is to be found is also 2. Substitute m = 2, 1 3x , and 1 2y into the point-slope form.
1 1 – –
2 2 – 3 2 2 – 6–2 –8 or 2 – 8
y y m x xy x y x
x y x y
(b) Solve for y. y = 2x – 8
53. (a) through (1, 6), perpendicular to 3x + 5y = 1 Find the slope of the line 3x + 5y = 1 by writing this equation in slope-intercept form.
3 15 5
3 5 1 5 –3 1–
x y y xy x
This line has a slope of 35
. The slope of
any line perpendicular to this line is 53
,
because 3 55 3
1. Substitute 53
m ,
1 1,x and 1 6y into the point-slope form.
53
– 6 ( –1)3( – 6) 5( –1)3 – 18 5 – 5
–13 5 – 3 or 5 – 3 –13
y xy xy x
x y x y
(b) Solve for y. 5 133 3
3 5 13y x y x
54. (a) through (–2, 0), perpendicular to 8x – 3y = 7 Find the slope of the line 8x – 3y = 7 by writing the equation in slope-intercept form.
8 73 3
8 – 3 7 –3 –8 7–
x y y xy x
This line has a slope of 83
. The slope of
any line perpendicular to this line is 38
,
because 8 33 8
1.
Substitute 38
m , 1 2,x and 1 0y
into the point-slope form. 38
– 0 – ( 2)8 –3( 2)8 –3 – 6 3 8 –6
y xy xy x x y
(b) Solve for y. 3 68 8
3 38 4
8 –3 – 6 –
–
y x y xy x
55. (a) through (4, 1), parallel to y = −5 Because y = −5 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, 1), the equation is y = 1.
(b) The slope-intercept form is y = 1.
56. (a) through 2, 2 , parallel to y = 3.
Because y = 3 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through 2, 2 ,
the equation is y = –2.
(b) The slope-intercept form is y = −2
57. (a) through (–5, 6), perpendicular to x = –2. Because x = –2 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (–5, 6), the equation is y = 6.
Section 2.5 Equations of Lines and Linear Models 209
Copyright © 2017 Pearson Education, Inc.
(b) The slope-intercept form is y = 6.
58. (a) Through (4, –4), perpendicular to x = 4 Because x = 4 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, –4), the equation is y = –4.
(b) The slope-intercept form is y = –4.
59. (a) Find the slope of the line 3y + 2x = 6.
23
3 2 6 3 –2 6– 2
y x y xy x
Thus, 23
– .m A line parallel to
3y + 2x = 6 also has slope 23
– .
Solve for k using the slope formula.
12
2 1 2–
4 33 2
–4 3
3 23 4 3 4 –
4 39 2 49 2 8
2 1
k
k
k kk
kk
k k
(b) Find the slope of the line 2y – 5x = 1.
5 12 2
2 5 1 2 5 1y x y xy x
Thus, 52
.m A line perpendicular to 2y
– 5x = 1 will have slope 25
– , because
5 22 5
– –1.
Solve this equation for k.
72
3 2
4 53 2
5 4 5 44 515 2 415 2 82 7
k
k kk
kk
k k
60. (a) Find the slope of the line 2 – 3 4.x y
2 43 3
2 – 3 4 –3 –2 4–
x y y xy x
Thus, 23
.m A line parallel to
2x – 3y = 4 also has slope 23
. Solve for r
using the slope formula. – 6 2 – 6 2
– 4 – 2 3 – 6 3
– 6 26 6
– 6 36 4 2
r r
r
r r
(b) Find the slope of the line x + 2y = 1.
1 12 2
2 1 2 – 1–
x y y xy x
Thus, 12
.m A line perpendicular to
the line x + 2y = 1 has slope 2, because 12
– (2) –1. Solve for r using the slope
formula. – 6 – 6
2 2– 4 – 2 – 6
– 6 –12 – 6
r r
r r
61. (a) First find the slope using the points (0, 6312) and (3, 7703).
7703 6312 1391463.67
3 0 3m
The y-intercept is (0, 6312), so the equation of the line is
463.67 6312.y x
(b) The value x = 4 corresponds to the year 2013.
463.67 4 6312 8166.68y
The model predicts that average tuition and fees were $8166.68 in 2013. This is $96.68 more than the actual amount.
62. (a) First find the slope using the points (0, 6312) and (2, 7136).
7136 6312 824412
2 0 2m
The y-intercept is (0, 6312), so the equation of the line is 412 6312.y x
(b) The value x = 4 corresponds to the year 2013.
412 4 6312 7960y
The model predicts that average tuition and fees were $7960 in 2013. This is $110 less than the actual amount.
210 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
63. (a) First find the slope using the points (0, 22036) and (4, 24525).
24525 22036 2489622.25
4 0 4m
The y-intercept is (0, 22036), so the equation of the line is
622.25 22,036.y x
The slope of the line indicates that the
average tuition increase is about $622 per year from 2009 through 2013.
(b) The year 2012 corresponds to x = 3. 622.25 3 22,036 23,902.75y
According to the model, average tuition and fees were $23,903 in 2012. This is $443 more than the actual amount $23,460.
(c) Using the linear regression feature, the equation of the line of best fit is
653 21,634.y x
64. (a) See the graph in the answer to part (b).There appears to be a linear relationship between the data. The farther the galaxy is from Earth, the faster it is receding.
(b) Using the points (520, 40,000) and (0, 0), we obtain
40,000 – 0 40,00076.9.
520 – 0 520m
The equation of the line through these two points is y = 76.9x.
(c) 76.9 60,00060,000
78076.9
x
x x
According to the model, the galaxy Hydra is approximately 780 megaparsecs away.
(d) 11
1110 9
9.5 10
9.5 101.235 10 12.35 10
76.9
Am
A
Using m = 76.9, we estimate that the age of the universe is approximately 12.35 billion years.
(e) 11
10 9
119
9.5 101.9 10 or 19 10
509.5 10
9.5 10100
A
A
The range for the age of the universe is between 9.5 billion and 19 billion years.
65. (a) The ordered pairs are (0, 32) and (100, 212).
The slope is 212 – 32 180 9
.100 – 0 100 5
m
Use 91 1 5
( , ) (0,32) and x y m in the
point-slope form.
1 195959 95 5
– ( – )
– 32 ( – 0)
– 32
32 32
y y m x xy xy x
y x F C
(b)
95
59
32
5 9 325 9 160 9 5 –1609 5( – 32) ( – 32)
F CF CF C C FC F C F
(c) 59
( – 32)9 5( – 32) 9 5 – 1604 –160 – 40
F C F FF F F FF F
F = C when F is –40º.
Section 2.5 Equations of Lines and Linear Models 211
Copyright © 2017 Pearson Education, Inc.
66. (a) The ordered pairs are (0, 1) and (100, 3.92). The slope is
3.92 – 1 2.920.0292
100 – 0 100m and 1.b
Using slope-intercept form we have 0.0292 1y x or ( ) 0.0292 1.p x x
(b) Let x = 60. P(60) = 0.0292(60) + 1 = 2.752 The pressure at 60 feet is approximately 2.75 atmospheres.
67. (a) Because we want to find C as a function of I, use the points (12026, 10089) and (14167, 11484), where the first component represents the independent variable, I. First find the slope of the line.
11484 10089 13950.6516
14167 12026 2141m
Now use either point, say (12026, 10089), and the point-slope form to find the equation.
–10089 0.6516( –12026)–10089 0.6516 – 7836
0.6516 2253
C IC I
C I
(b) Because the slope is 0.6516, the marginal propensity to consume is 0.6516.
68. D is the only possible answer, because the x-intercept occurs when y = 0. We can see from the graph that the value of the x-intercept exceeds 10.
69. Write the equation as an equivalent equation with 0 on one side: 2 7 4 2x x x 2 7 4 2 0x x x . Now graph
2 7 4 2y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:
Solution set: {3}
70. Write the equation as an equivalent equation with 0 on one side: 7 2 4 5 3 1x x x 7 2 4 5 3 1 0x x x . Now graph
7 2 4 5 3 1y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:
Solution set: {1}
71. Write the equation as an equivalent equation with 0 on one side: 3 2 1 2 2 5x x
3 2 1 2 2 5 0x x . Now graph
3 2 1 2 2 5y x x in the window
[–5, 5] × [–5, 5] to find the x-intercept:
Solution set: 12
or 0.5
72. Write the equation as an equivalent equation with 0 on one side:
4 3 4 2 2 3 6 2x x x x
4 3 4 2 2 3 6 2 0x x x x .
Now graph 4 3 4 2 2 3 6 2y x x x x in the
window [–2, 8] × [–5, 5] to find the x-intercept:
Solution set: {4}
73. (a) 2 5 22 10 2
10 212
x xx x
xx
Solution set: {12}
212 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b) Answers will vary. Sample answer: The solution does not appear in the standard viewing window x-interval [10, –10]. The minimum and maximum values must include 12.
74. Rewrite the equation as an equivalent equation with 0 on one side.
3 2 6 4 8 2
6 18 4 8 2 0
x x xx x x
Now graph y = −6x − 18 − (−4x + 8 −2x) in the window [–10, 10] × [–30, 10].
The graph is a horizontal line that does not
intersect the x-axis. Therefore, the solution set is . We can verify this algebraically.
3 2 6 4 8 26 18 6 8 0 26
x x xx x
Because this is a false statement, the solution set is .
75. A(–1, 4), B(–2, –1), C(1, 14)
For A and B, 1 4 5
52 ( 1) 1
m
For B and C, 14 ( 1) 15
51 ( 2) 3
m
For A and C, 14 4 10
51 ( 1) 2
m
Since all three slopes are the same, the points are collinear.
76. A(0, −7), B(–3, 5), C(2, −15)
For A and B, 5 ( 7) 12
43 0 3
m
For B and C, 15 5 20
42 ( 3) 5
m
For A and C, 15 ( 7) 8
42 0 2
m
Since all three slopes are the same, the points are collinear.
77. A(−1, −3), B(−5, 12), C(1, −11)
For A and B, 12 ( 3) 15
5 ( 1) 4m
For B and C, 11 12 23
1 ( 5) 6m
For A and C, 11 ( 3) 8
41 ( 1) 2
m
Since all three slopes are not the same, the points are not collinear.
78. A(0, 9), B(–3, –7), C(2, 19)
For A and B, 7 9 16 16
3 0 3 3m
For B and C, 19 ( 7) 26
2 ( 3) 5m
For A and C, 19 9 10
52 0 2
m
Because all three slopes are not the same, the points are not collinear.
79. 2 21 1 1
2 2 21 1 1
( , ) ( – 0) ( – 0)d O P x m x
x m x
80. 2 22 2 2
2 2 22 2 2
( , ) ( – 0) ( – 0)d O Q x m x
x m x
81. 2 22 1 2 2 1 1( , ) ( – ) ( – )d P Q x x m x m x
82.
2 2 2
2 22 2 2 2 2 2
1 1 1 2 2 2
22 2
2 1 2 2 1 1
2 2 2 2 2 21 1 1 2 2 2
2 22 1 2 2 1 1
[ ( , )] [ ( , )] [ ( , )]
– –
– –
d O P d O Q d P Q
x m x x m x
x x m x m x
x m x x m x
x x m x m x
2 2 2 2 2 21 1 1 2 2 2
2 2 2 22 2 1 1 2 2
2 21 2 1 2 1 1
2 1 1 2 1 2
1 2 1 2 1
2
20 2 2
2 2 0
x m x x m xx x x x m x
m m x x m xx x m m x x
m m x x x x
83. 1 2 1 2 1 2
1 2 1 2
–2 – 2 0–2 ( 1) 0m m x x x x
x x m m
84. 1 2 1 2–2 ( 1) 0x x m m
Because 1 20 and 0, x x we have
1 2 1 21 0 implying that –1.m m m m
Summary Exercises on Graphs, Circles, Functions, and Equations 213
Copyright © 2017 Pearson Education, Inc.
85. If two nonvertical lines are perpendicular, then the product of the slopes of these lines is –1.
Summary Exercises on Graphs, Circles, Functions, and Equations
1. (3, 5), (2, 3)P Q
(a)
2 2
2 2
( , ) (2 3) ( 3 5)
1 8
1 64 65
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
5 33 2 5 2 5, , ,1
2 2 2 2 2
.
(c) First find m: 3 – 5 8
82 – 3 1
m
Use either point and the point-slope form. – 5 8 – 3y x
Change to slope-intercept form. 5 8 24 8 19y x y x
2. P(–1, 0), Q(4, –2)
(a) 2 2
2 2
( , ) [4 – (–1)] (–2 – 0)
5 (–2)
25 4 29
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
–1 4 0 (–2) 3 2, ,
2 2 2 23
, 1 .2
(c) First find m: 2 – 0 2 2
4 – 1 5 5m
Use either point and the point-slope form.
25
– 0 – 1y x
Change to slope-intercept form.
2 25 5
5 2 15 2 2
y xy xy x
3. P(–2, 2), Q(3, 2)
(a) 22
2 2
( , ) [3 – (– 2)] 2 2
5 0 25 0 25 5
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
– 2 3 2 2,
2 2
1 4 1, , 2 .
2 2 2
(c) First find m: 2 – 2 0
03 – 2 5
m
All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes through (3, 2), the equation is y = 2.
4. 2 2, 2 ,P 2,3 2Q
(a)
2 2
2 2
( , ) 2 – 2 2 3 2 – 2
– 2 2 2
2 8 10
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
2 2 2 2 3 2,
2 2
3 2 4 2 3 2, , 2 2 .
2 2 2
(c) First find m: 3 2 2 2 2
22 2 2 2
m
Use either point and the point-slope form.
– 2 2 – 2 2y x
Change to slope-intercept form.
– 2 2 4 2 2 5 2y x y x
5. P(5, –1), Q(5, 1)
(a) 2 2
2 2
( , ) (5 – 5) [1 – (–1)]
0 2 0 4 4 2
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
5 5 –1 1,
2 2
10 0
, 5,0 .2 2
214 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(c) First find m. 1 – 1 2
undefined5 – 5 0
m
All lines that have an undefined slope are vertical lines. The equation of a vertical line has an equation of the form x = a. The line passes through (5, 1), so the equation is x = 5 . (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)
6. (1, 1), ( 3, 3)P Q
(a)
2 2
2 2
( , ) ( 3 – 1) ( 3 –1)
4 4
16 16 32 4 2
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
1 3 1 3,
2 2
2 2,
2 2
1, 1 .
(c) First find m: 3 –1 4
13 –1 4
m
Use either point and the point-slope form. –1 1 –1y x
Change to slope-intercept form. 1 1y x y x
7. 2 3,3 5 , 6 3,3 5P Q
(a)
2 2
2 2
( , ) 6 3 – 2 3 3 5 – 3 5
4 3 0 48 4 3
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
2 3 6 3 3 5 3 5,
2 2
8 3 6 5, 4 3, 3 5 .
2 2
(c) First find m: 3 5 3 5 0
06 3 2 3 4 3
m
All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes
through 2 3,3 5 , the equation is
3 5.y
8. P(0, –4), Q(3, 1)
(a) 2 2
2 2
( , ) (3 – 0) [1 – (– 4)]
3 5 9 25 34
d P Q
(b) The midpoint M of the segment joining points P and Q has coordinates
0 3 – 4 1 3 3 3 3, , , .
2 2 2 2 2 2
(c) First find m: 1 – 4 5
3 0 3m
Using slope-intercept form we have 53
– 4.y x
9. Through 2,1 and 4, 1
First find m: 1 –1 2 1
4 – (–2) 6 3m
Use either point and the point-slope form.
13
– 1 – 4y x
Change to slope-intercept form.
1 13 3
3 1 4 3 3 4
3 1
y x y xy x y x
Summary Exercises on Graphs, Circles, Functions, and Equations 215
Copyright © 2017 Pearson Education, Inc.
10. the horizontal line through (2, 3) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (2, 3), the equation is y = 3.
11. the circle with center (2, –1) and radius 3
22 2
2 2
( – 2) – (–1) 3
( – 2) ( 1) 9
x yx y
12. the circle with center (0, 2) and tangent to the x-axis The distance from the center of the circle to the x-axis is 2, so r = 2.
2 2 2 2 2( – 0) ( – 2) 2 ( – 2) 4x y x y
13. the line through (3, 5) with slope 56
Write the equation in point-slope form.
56
– 5 3 y x
Change to standard form.
5 156 6
5 56 2
6 5 5 3 6 30 5 15
6 5 15
y x y xy x y x
y x
14. the vertical line through (−4, 3) The equation of a vertical line has an equation of the form x = a. Because the line passes through (−4, 3), the equation is x = −4.
15. a line through (–3, 2) and parallel to the line 2x + 3y = 6 First, find the slope of the line 2x + 3y = 6 by writing this equation in slope-intercept form.
23
2 3 6 3 2 6 2x y y x y x
The slope is 23
. Because the lines are
parallel, 23
is also the slope of the line
whose equation is to be found. Substitute 23
m , 1 3x , and 1 2y into the point-
slope form.
21 1 3
23
– – 2 3
3 2 2 3 3 6 2 – 6
3 –2
y y m x x y xy x y x
y x y x
16. a line through the origin and perpendicular to the line 3 4 2x y
First, find the slope of the line 3 4 2x y by writing this equation in slope-intercept form.
3 32 14 4 4 2
3 4 2 4 –3 2x y y xy x y x
This line has a slope of 34
. The slope of any
line perpendicular to this line is 43
, because
343 4
1. Using slope-intercept form we
have 4 43 3
0 or .y x y x
(continued on next page)
216 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
17. 2 24 2 4x x y y Complete the square on x and y separately.
2 2
2 2
2 2
4 2 4
4 4 2 1 4 4 1
2 1 9
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (2, −1) and radius 3.
18. 2 26 10 36 0x x y y Complete the square on x and y separately.
2 2
2 2
2 2
6 10 36
6 9 10 25 –36 9 25
3 5 2
x x y y
x x y y
x y
No, it is not a circle.
19. 2 212 20 0x x y Complete the square on x and y separately.
2 2
2 2
2 2
12 20
12 36 –20 36
6 16
x x y
x x y
x y
Yes, it is a circle. The circle has its center at (6, 0) and radius 4.
20. 2 22 16 61x x y y Complete the square on x and y separately.
2 2
2 2
2 2
2 16 61
2 1 16 64 –61 1 64
1 8 4
x x y y
x x y y
x y
Yes, it is a circle. The circle has its center at (−1, −8) and radius 2.
21. 2 22 10 0x x y Complete the square on x and y separately.
2 2
2 2
2 2
2 10
2 1 –10 1
1 9
x x y
x x y
x y
No, it is not a circle.
22. 2 2 – 8 9 0x y y Complete the square on x and y separately.
2 2
2 2
22
– 8 9
– 8 16 9 16
– 4 25
x y y
x y y
x y
Yes, it is a circle. The circle has its center at (0, 4) and radius 5.
23. The equation of the circle is 2 2 2( 4) ( 5) 4 .x y
Let y = 2 and solve for x: 2 2 2( 4) (2 5) 4x 2 2 2 2( 4) ( 3) 4 ( 4) 7x x
4 7 4 7x x
The points of intersection are 4 7, 2 and
4 7, 2
24. Write the equation in center-radius form by completing the square on x and y separately:
2 2
2 2
2 2
2 2
10 24 144 0
10 24 144 0
( 10 25) ( 24 144) 25
( 5) ( 12) 25
x y x yx x y y
x x y yx y
The center of the circle is (5, 12) and the radius is 5. Now use the distance formula to find the distance from the center (5, 12) to the origin:
2 22 1 2 1
2 2(5 0) (12 0) 25 144 13
d x x y y
The radius is 5, so the shortest distance from the origin to the graph of the circle is 13 − 5 = 8.
(continued on next page)
Section 2.6 Graphs of Basic Functions 217
Copyright © 2017 Pearson Education, Inc.
(continued)
25. (a) The equation can be rewritten as 6 31 1
4 4 4 24 6 .y x y x y x
x can be any real number, so the domain is all real numbers and the range is also all real numbers. domain: , ; range: ,
(b) Each value of x corresponds to just one value of y. 4 6x y represents a function.
3 31 14 2 4 2
y x f x x
3 31 1 24 2 2 2 2
2 2 1f
26. (a) The equation can be rewritten as 2 5 .y x y can be any real number.
Because the square of any real number is
not negative, 2y is never negative. Taking the constant term into consideration, domain would be 5, .
domain: 5, ; range: ,
(b) Because (−4, 1) and (−4, −1) both satisfy
the relation, 2 5y x does not represent a function.
27. (a) 2 22 25x y is a circle centered at
(−2, 0) with a radius of 5. The domain will start 5 units to the left of –2 and end 5 units to the right of –2. The domain will be [−2 − 5, 2 + 5] = [−7, 3]. The range will start 5 units below 0 and end 5 units above 0. The range will be [0 − 5, 0 + 5] = [−5, 5].
(b) Because (−2, 5) and (−2, −5) both satisfy
the relation, 2 22 25x y does not
represent a function.
28. (a) The equation can be rewritten as 2 2 31
2 22 3 .y x y x x can be
any real number. Because the square of
any real number is not negative, 212
x is
never negative. Taking the constant term into consideration, range would be
32
, .
domain: , ; range: 32
,
(b) Each value of x corresponds to just one
value of y. 2 2 3x y represents a function.
2 23 31 12 2 2 2
y x f x x
2 3 3 31 1 4 12 2 2 2 2 2 2
2 2 4f
Section 2.6 Graphs of Basic Functions
1. The equation 2f x x matches graph E.
The domain is , .
2. The equation of f x x matches graph G.
The function is increasing on 0, .
3. The equation 3f x x matches graph A.
The range is , .
4. Graph C is not the graph of a function.
Its equation is 2x y .
5. Graph F is the graph of the identity function. Its equation is .f x x
6. The equation f x x matches graph B.
1.5 1f
7. The equation 3f x x matches graph H.
No, there is no interval over which the function is decreasing.
8. The equation of f x x matches graph D.
The domain is 0, .
9. The graph in B is discontinuous at many points. Assuming the graph continues, the range would be {…, −3, −2, −1, 0, 1, 2, 3, …}.
218 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
10. The graphs in E and G decrease over part of the domain and increase over part of the domain. They both increase over 0, and
decrease over , 0 .
11. The function is continuous over the entire domain of real numbers , .
12. The function is continuous over the entire domain of real numbers , .
13. The function is continuous over the interval
0, .
14. The function is continuous over the interval , 0 .
15. The function has a point of discontinuity at (3, 1). It is continuous over the interval ,3 and the interval 3, .
16. The function has a point of discontinuity at x = 1. It is continuous over the interval ,1
and the interval 1, .
17. 2 if –1( )
–1 if –1x xf x x x
(a) (–5) 2(–5) –10f
(b) (–1) 2(–1) –2f
(c) f(0) = 0 –1 = –1
(d) f(3) = 3 – 1 = 2
18. – 2 if 3( )
5 – if 3x xf x x x
(a) f(–5) = –5 – 2 = –7
(b) f(–1) = –1 – 2 = –3
(c) f(0) = 0 – 2 = –2
(d) f(3) = 5 – 3 = 2
19. 2 if – 4
( ) – if – 4 23 if 2
x xf x x x
x x
(a) f(–5) = 2 + (–5) = –3
(b) f(–1) = – (–1) = 1
(c) f(0) = –0 = 0
(d) (3) 3 3 9f
20. –2 if –33 –1 if – 3 2– 4 if 2
x xf x x x
x x
(a) f(–5) = –2(–5) = 10
(b) f(–1) = 3(–1) – 1 = –3 – 1 = –4
(c) f(0) = 3(0) – 1 = 0 – 1 = –1
(d) f(3) = –4(3) = –12
21. – 1 if 3( )
2 if 3x xf x x
Draw the graph of y = x – 1 to the left of x = 3, including the endpoint at x = 3. Draw the graph of y = 2 to the right of x = 3, and note that the endpoint at x = 3 coincides with the endpoint of the other ray.
22. 6 – if 3( )
3 if 3x xf x x
Graph the line y = 6 – x to the left of x = 3, including the endpoint. Draw y = 3 to the right of x = 3. Note that the endpoint at x = 3 coincides with the endpoint of the other ray.
23. 4 – if 2( )
1 2 if 2x xf x x x
Draw the graph of y = 4 − x to the left of x = 2, but do not include the endpoint. Draw the graph of y = 1 + 2x to the right of x = 2, including the endpoint.
Section 2.6 Graphs of Basic Functions 219
Copyright © 2017 Pearson Education, Inc.
24. 2 1 if 0( )
if 0x xf x x x
Graph the line y = 2x + 1 to the right of x = 0, including the endpoint. Draw y = x to the left of x = 0, but do not include the endpoint.
25. 3 if 1( )
1 if 1xf x x
Graph the line y = −3 to the left of x = 1, including the endpoint. Draw y = −1 to the right of x = 1, but do not include the endpoint.
26. 2 if 1( )
2 if 1xf x x
Graph the line y = – 2 to the left of x = 1, including the endpoint. Draw y = 2 to the right of x = 1, but do not include the endpoint.
Exercise 26 Exercise 27
27. 2 if – 4
( ) – if – 4 53 if 5
x xf x x x
x x
Draw the graph of y = 2 + x to the left of –4, but do not include the endpoint at x = 4. Draw the graph of y = –x between –4 and 5, including both endpoints. Draw the graph of y = 3x to the right of 5, but do not include the endpoint at x = 5.
28. –2 if –3
( ) 3 –1 if – 3 2– 4 if 2
x xf x x x
x x
Graph the line y = –2x to the left of x = –3, but do not include the endpoint. Draw y = 3x – 1 between x = –3 and x = 2, and include both endpoints. Draw y = –4x to the right of x = 2, but do not include the endpoint. Notice that the endpoints of the pieces do not coincide.
29. 21
212
– +2 if 2( )
if 2
x xf x
x x
Graph the curve 212
2y x to the left of
x = 2, including the endpoint at (2, 0). Graph
the line 12
y x to the right of x = 2, but do
not include the endpoint at (2, 1). Notice that the endpoints of the pieces do not coincide.
30. 3
25 if 0
( )if 0
x xf xx x
Graph the curve 3 5y x to the left of x = 0, including the endpoint at (0, 5). Graph
the line 2y x to the right of x = 0, but do not include the endpoint at (0, 0). Notice that the endpoints of the pieces do not coincide.
220 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
31. 2
2 if 5 1( ) 2 if 1 0
2 if 0 2
x xf x x
x x
Graph the line 2y x between x = −5 and x = −1, including the left endpoint at (−5, −10), but not including the right endpoint at (−1, −2). Graph the line y = −2 between x = −1 and x = 0, including the left endpoint at (−1, −2) and not including the right endpoint at (0, −2). Note that (−1, −2) coincides with the first two sections, so it is included. Graph
the curve 2 2y x from x = 0 to x = 2, including the endpoints at (0, −2) and (2, 2).
Note that (0, −2) coincides with the second two sections, so it is included. The graph ends at x = −5 and x = 2.
32.
2
2
0.5 if 4 2( ) if 2 2
4 if 2 4
x xf x x x
x x
Graph the curve 20.5y x between x = −4 and x = −2, including the endpoints at (−4, 8) and (−2, 2). Graph the line y x between x = −2 and x = 2, but do not include the endpoints at (−2, −2) and (2, 2). Graph the
curve 2 4y x from x = 2 to x = 4, including the endpoints at (2, 0) and (4, 12). The graph ends at x = −4 and x = 4.
33.
3
2
3 if 2 0( ) 3 if 0 1
4 if 1 3
x xf x x x
x x x
Graph the curve 3 3y x between x = −2 and x = 0, including the endpoints at (−2, −5) and (0, 3). Graph the line y = x + 3 between x = 0 and x = 1, but do not include the endpoints at (0, 3) and (1, 4). Graph the curve
24y x x from x = 1 to x = 3, including the endpoints at (1, 4) and (3, −2). The graph ends at x = −2 and x = 3.
34. 2
312
2 if 3 1
( ) 1 if 1 2
1 if 2 3
x xf x x x
x x
Graph the curve y = −2x to from x = −3 to x = −1, including the endpoint (−3, 6), but not including the endpoint (−1, 2). Graph the
curve 2 1y x from x = −1 to x = 2, including the endpoints (−1, 2) and (2, 5).
Graph the curve 312
1y x from x = 2 to
x = 3, including the endpoint (3, 14.5) but not including the endpoint (2, 5). Because the endpoints that are not included coincide with endpoints that are included, we use closed dots on the graph.
Section 2.6 Graphs of Basic Functions 221
Copyright © 2017 Pearson Education, Inc.
35. The solid circle on the graph shows that the endpoint (0, –1) is part of the graph, while the open circle shows that the endpoint (0, 1) is not part of the graph. The graph is made up of parts of two horizontal lines. The function which fits this graph is
–1 if 0( )
1 if 0.xf x x
domain: , ; range: {–1, 1}
36. We see that y = 1 for every value of x except x = 0, and that when x = 0, y = 0. We can write the function as
1 if 0( )
0 if 0.xf x x
domain: , ; range: {0, 1}
37. The graph is made up of parts of two horizontal lines. The solid circle shows that the endpoint (0, 2) of the one on the left belongs to the graph, while the open circle shows that the endpoint (0, –1) of the one on the right does not belong to the graph. The function that fits this graph is
2 if 0( )
–1 if 1.xf x x
domain: ( , 0] (1, ); range: {–1, 2}
38. We see that y = 1 when 1x and that y = –1 when x > 2. We can write the function as
1 if –1( )
–1 if 2.xf x x
domain: (– , –1] (2, ); range: {–1, 1}
39. For 0x , that piece of the graph goes through the points (−1, −1) and (0, 0). The slope is 1, so the equation of this piece is y = x. For x > 0, that piece of the graph is a horizontal line passing through (2, 2), so its equation is y = 2. We can write the function as
if 0( ) .
2 if 0x xf x x
domain: (– , ) range: ( , 0] {2}
40. For x < 0, that piece of the graph is a horizontal line passing though (−3, −3), so the equation of this piece is y = −3. For 0x , the curve passes through (1, 1) and (4, 2), so the
equation of this piece is y x . We can
write the function as 3 if 0
( ) . if 0
xf x
x x
domain: (– , ) range: { 3} [0, )
41. For x < 1, that piece of the graph is a curve passes through (−8, −2), (−1, −1) and (1, 1), so
the equation of this piece is 3y x . The right piece of the graph passes through (1, 2) and
(2, 3). 2 3
11 2
m
, and the equation of the
line is 2 1 1y x y x . We can write
the function as 3 if 1( )
1 if 1x xf x
x x
domain: (– , ) range: ( ,1) [2, )
42. For all values except x = 2, the graph is a line. It passes through (0, −3) and (1, −1). The slope is 2, so the equation is y = 2x −3. At x = 2, the graph is the point (2, 3). We can write
the function as 3 if 2( ) .
2 3 if 2xf x x x
domain: (– , ) range: ( ,1) (1, )
43. f(x) = x
Plot points.
x –x f(x) = x
–2 2 2
–1.5 1.5 1
–1 1 1
–0.5 0.5 0
0 0 0
0.5 –0.5 –1
1 –1 –1
1.5 –1.5 –2
2 –2 –2
More generally, to get y = 0, we need 0 – 1 0 1 1 0.x x x To get y = 1, we need 1 – 2x
1 2 –2 1.x x Follow this pattern to graph the step function.
domain: , ; range: {...,–2,–1,0,1,2,...}
222 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
44. f(x) = x
Plot points.
x x f(x) = x
–2 −2 2
–1.5 −2 2
–1 −1 1
–0.5 −1 1
0 0 0
0.5 0 0
1 1 −1
1.5 1 –1
2 2 –2
Follow this pattern to graph the step function.
domain: , ; range: {...,–2,–1,0,1,2,...}
45. f(x) = 2x
To get y = 0, we need 12
0 2 1 0 .x x
To get y = 1, we need 12
1 2 2 1.x x
To get y = 2, we need 32
2 2 3 1 .x x
Follow this pattern to graph the step function.
domain: , ; range: {...,–2,–1,0,1,2,...}
46. g(x) = 2 1x
To get y = 0, we need 12
0 2 1 1 1 2 2 1.x x x
To get y = 1, we need 32
1 2 – 1 2 2 2 3 1 .x x x
Follow this pattern to graph the step function.
domain: , ; range: {..., 2,–1,0,1,2,...}
47. The cost of mailing a letter that weighs more than 1 ounce and less than 2 ounces is the same as the cost of a 2-ounce letter, and the cost of mailing a letter that weighs more than 2 ounces and less than 3 ounces is the same as the cost of a 3-ounce letter, etc.
48. The cost is the same for all cars parking
between 12
hour and 1-hour, between 1 hour
and 12
1 hours, etc.
49.
Section 2.7 Graphing Techniques 223
Copyright © 2017 Pearson Education, Inc.
50.
51. (a) For 0 8,x 49.8 34.2
1.958 0
m
,
so 1.95 34.2.y x For 8 13x ,
52.2 49.80.48
13 8m
, so the equation
is 52.2 0.48( 13)y x 0.48 45.96y x
(b) 1.95 34.2 if 0 8( )
0.48 45.96 if 8 13x xf x x x
52. When 0 3x , the slope is 5, which means that the inlet pipe is open, and the outlet pipe is closed. When 3 5x , the slope is 2, which means that both pipes are open. When 5 8x , the slope is 0, which means that both pipes are closed. When 8 10x , the slope is −3, which means that the inlet pipe is closed, and the outlet pipe is open.
53. (a) The initial amount is 50,000 gallons. The final amount is 30,000 gallons.
(b) The amount of water in the pool remained constant during the first and fourth days.
(c) (2) 45,000; (4) 40,000f f
(d) The slope of the segment between (1, 50000) and (3, 40000) is −5000, so the water was being drained at 5000 gallons per day.
54. (a) There were 20 gallons of gas in the tank at x = 3.
(b) The slope is steepest between t = 1 and t ≈ 2.9, so that is when the car burned gasoline at the fastest rate.
55. (a) There is no charge for additional length, so we use the greatest integer function. The cost is based on multiples of two
feet, so 2
( ) 0.8 xf x if 6 18x .
(b) 8.52
(8.5) 0.8 0.8(4) $3.20f
15.22
(15.2) 0.8 0.8(7) $5.60f
56. (a) 6.5 if 0 4
( ) 5.5 48 if 4 6–30 195 if 6 6.5
x xf x x x
x x
Draw a graph of y = 6.5x between 0 and 4, including the endpoints. Draw the graph of y = –5.5x + 48 between 4 and 6, including the endpoint at 6 but not the one at 4. Draw the graph of y = –30x + 195, including the endpoint at 6.5 but not the one at 6. Notice that the endpoints of the three pieces coincide.
(b) From the graph, observe that the snow depth, y, reaches its deepest level (26 in.) when x = 4, x = 4 represents 4 months after the beginning of October, which is the beginning of February.
(c) From the graph, the snow depth y is nonzero when x is between 0 and 6.5. Snow begins at the beginning of October and ends 6.5 months later, in the middle of April.
Section 2.7 Graphing Techniques
1. To graph the function 2 3,f x x shift the
graph of 2y x down 3 units.
2. To graph the function 2 5,f x x shift the
graph of 2y x up 5 units.
3. The graph of 24f x x is obtained by
shifting the graph of 2y x to the left 4 units.
4. The graph of 27f x x is obtained by
shifting the graph of 2y x to the right 7 units.
5. The graph of f x x is a reflection of
the graph of f x x across the x-axis.
6. The graph of f x x is a reflection of
the graph of f x x across the y-axis.
224 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
7. To obtain the graph of 32 3,f x x
shift the graph of 3y x 2 units to the left and 3 units down.
8. To obtain the graph of 33 6,f x x
shift the graph of 3y x 3 units to the right and 6 units up.
9. The graph of f x x is the same as the
graph of y x because reflecting it across
the y-axis yields the same ordered pairs.
10. The graph of 2x y is the same as the graph
of 2x y because reflecting it across the
x-axis yields the same ordered pairs.
11. (a) B; 2( 7)y x is a shift of 2 ,y x 7 units to the right.
(b) D; 2 7y x is a shift of 2 ,y x 7 units downward.
(c) E; 27y x is a vertical stretch of 2 ,y x by a factor of 7.
(d) A; 2( 7)y x is a shift of 2 ,y x 7 units to the left.
(e) C; 2 7y x is a shift of 2 ,y x 7 units upward.
12. (a) E; 34y x is a vertical stretch
of 3 ,y x by a factor of 4.
(b) C; 3y x is a reflection of 3 ,y x over the x-axis.
(c) D; 3y x is a reflection of 3 ,y x over the y-axis.
(d) A; 3 4y x is a shift of 3 ,y x 4 units to the right.
(e) B; 3 4y x is a shift of 3 ,y x 4 units down.
13. (a) B; 2 2y x is a shift of 2 ,y x 2 units upward.
(b) A; 2 2y x is a shift of 2 ,y x 2 units downward.
(c) G; 2( 2)y x is a shift of 2 ,y x 2 units to the left.
(d) C; 2( 2)y x is a shift of 2 ,y x 2 units to the right.
(e) F; 22y x is a vertical stretch of 2 ,y x by a factor of 2.
(f) D; 2y x is a reflection of 2 ,y x across the x-axis.
(g) H; 2( 2) 1y x is a shift of 2 ,y x 2 units to the right and 1 unit upward.
(h) E; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit upward.
(i) I; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit down.
14. (a) G; 3y x is a shift of ,y x 3 units to the left.
(b) D; 3y x is a shift of ,y x 3 units downward.
(c) E; 3y x is a shift of ,y x 3 units upward.
(d) B; 3y x is a vertical stretch of
,y x by a factor of 3.
(e) C; y x is a reflection of y x across the x-axis.
(f) A; 3y x is a shift of ,y x 3 units to the right.
(g) H; 3 2y x is a shift of ,y x 3 units to the right and 2 units upward.
(h) F; 3 2y x is a shift of ,y x 3 units to the left and 2 units upward.
(i) I; 3 2y x is a shift of ,y x 3 units to the right and 2 units downward.
15. (a) F; 2y x is a shift of y x 2 units
to the right.
(b) C; 2y x is a shift of y x 2 units
downward.
(c) H; 2y x is a shift of y x 2 units
upward.
Section 2.7 Graphing Techniques 225
Copyright © 2017 Pearson Education, Inc.
(d) D; 2y x is a vertical stretch of y x
by a factor of 2.
(e) G; y x is a reflection of
y x across the x-axis.
(f) A; y x is a reflection of y x
across the y-axis.
(g) E; 2y x is a reflection of 2y x
across the x-axis. 2y x is a vertical
stretch of y x by a factor of 2.
(h) I; 2 2y x is a shift of y x 2
units to the right and 2 units upward.
(i) B; 2 2y x is a shift of y x 2
units to the left and 2 units downward.
16. The graph of 32 1 6f x x is the graph
of 3f x x stretched vertically by a factor
of 2, shifted left 1 unit and down 6 units.
17. 3f x x
x h x x 3f x x
−2 2 6
−1 1 3
0 0 0
1 1 3
2 2 6
18. 4f x x
x h x x 4f x x
−2 2 8
−1 1 4
0 0 0
1 1 4
2 2 8
19. 23
f x x
x h x x 23
f x x
−3 3 2
−2 2 43
−1 1 23
0 0 0
1 1 23
2 2 43
3 3 2
20. 34
f x x
x h x x 34
f x x
−4 4 3
−3 3 94
−2 2 32
−1 1 34
0 0 0
1 1 34
2 2 32
3 3 94
4 4 3
(continued on next page)
226 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
21. 22f x x
x 2h x x 22f x x
−2 4 8
−1 1 2
0 0 0
1 1 2
2 4 8
22. 23f x x
x 2h x x 23f x x
−2 4 12
−1 1 3
0 0 0
1 1 3
2 4 12
23. 212
f x x
x 2h x x 212
f x x
−2 4 2
−1 1 12
0 0 0
1 1 12
2 4 2
24. 213
f x x
x 2h x x 213
f x x
−3 9 3
−2 4 43
−1 1 13
0 0 0
1 1 13
2 4 43
3 9 3
Section 2.7 Graphing Techniques 227
Copyright © 2017 Pearson Education, Inc.
25. 212
f x x
x 2h x x 212
f x x
−3 9 92
−2 4 2
−1 1 12
0 0 0
1 1 12
2 4 2
3 9 92
26. 213
f x x
x 2h x x 213
f x x
−3 9 −3
−2 4 43
−1 1 13
0 0 0
1 1 13
2 4 43
3 9 −3
27. 3f x x
x h x x 3f x x
−2 2 −6
−1 1 −3
0 0 0
1 1 −3
2 2 −6
28. 2f x x
x h x x 2f x x
−2 2 −4
−1 1 −2
0 0 0
1 1 −2
2 2 −4
29. 12
h x x
x f x x 1
21 12 2
h x x
x x
−4 4 2
−3 3 32
−2 2 1
−1 1 12
0 0 0
(continued on next page)
228 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
x f x x 1
21 12 2
h x x
x x
1 1 12
2 2 1
3 3 32
4 4 2
30. 13
h x x
x 13
f x x 1
31 13 3
h x x
x x
−3 3 1
−2 2 23
−1 1 13
0 0 0
1 1 13
2 2 23
3 3 1
31. 4h x x
x f x x 4 2h x x x
0 0 0
1 1 2
2 2 2 2
x f x x 4 2h x x x
3 3 2 3
4 2 4
32. 9h x x
x f x x 9 3h x x x
0 0 0
1 1 3
2 2 3 2
3 3 3 3
4 2 6
33. f x x
x h x x f x x
−4 2 −2
−3 3 3
−2 2 2
−1 1 −1
0 0 0
Section 2.7 Graphing Techniques 229
Copyright © 2017 Pearson Education, Inc.
34. f x x
x h x x f x x
−3 3 −3
−2 2 −2
−1 1 −1
0 0 0
1 1 −1
2 2 −2
3 3 −3
35. (a) 4y f x is a horizontal translation
of f, 4 units to the left. The point that corresponds to (8, 12) on this translated function would be 8 4,12 4,12 .
(b) 4y f x is a vertical translation of f, 4 units up. The point that corresponds to (8, 12) on this translated function would be 8,12 4 8,16 .
36. (a) 14
y f x is a vertical shrinking of f, by
a factor of 14
. The point that corresponds
to (8, 12) on this translated function
would be 14
8, 12 8,3 .
(b) 4y f x is a vertical stretching of f, by
a factor of 4. The point that corresponds to (8, 12) on this translated function would be 8, 4 12 8, 48 .
37. (a) (4 )y f x is a horizontal shrinking of f, by a factor of 4. The point that corresponds to (8, 12) on this translated
function is 14
8 , 12 2, 12 .
(b) 14
y f x is a horizontal stretching of f, by a factor of 4. The point that corresponds to (8, 12) on this translated function is 8 4, 12 32, 12 .
38. (a) The point that corresponds to (8, 12) when reflected across the x-axis would be (8, −12).
(b) The point that corresponds to (8, 12) when reflected across the y-axis would be (−8, 12).
39. (a) The point that is symmetric to (5, –3) with respect to the x-axis is (5, 3).
(b) The point that is symmetric to (5, –3) with respect to the y-axis is (–5, –3).
(c) The point that is symmetric to (5, –3) with respect to the origin is (–5, 3).
40. (a) The point that is symmetric to (–6, 1) with respect to the x-axis is (–6, –1).
(b) The point that is symmetric to (–6, 1) with respect to the y-axis is (6, 1).
(c) The point that is symmetric to (–6, 1) with respect to the origin is (6, –1).
41. (a) The point that is symmetric to (–4, –2) with respect to the x-axis is (–4, 2).
(b) The point that is symmetric to (–4, –2) with respect to the y-axis is (4, –2).
(c) The point that is symmetric to (–4, –2) with respect to the origin is (4, 2).
230 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
42. (a) The point that is symmetric to (–8, 0) with respect to the x-axis is (–8, 0) because this point lies on the x-axis.
(b) The point that is symmetric to the point (–8, 0) with respect to the y-axis is (8, 0).
(c) The point that is symmetric to the point (–8, 0) with respect to the origin is (8, 0).
43. The graph of y = |x − 2| is symmetric with respect to the line x = 2.
44. The graph of y = −|x + 1| is symmetric with respect to the line x = −1.
45. 2 5y x Replace x with –x to obtain
2 2( ) 5 5y x x . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
2 2 2( ) 2 2 2.y x y x y x The result is not the same as the original
equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.
46. 42 3y x Replace x with –x to obtain
4 42( ) 3 2 3y x x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to
obtain 4 4– 2( ) 3 2 3y x y x 42 3y x . The result is not the same as
the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.
47. 2 2 12x y Replace x with –x to obtain
2 2 2 2( ) 12 12x y x y . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2 2 2( ) 12 12x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin.
48. 2 2 6y x Replace x with –x to obtain
22 2 26 6y x y x
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2 2 2( ) 6 6y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis, the y-axis, and the origin.
49. 34y x x Replace x with –x to obtain
3 3
3
4( ) ( ) 4( )
4 .
y x x y x xy x x
The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to
obtain 3 34 4 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
3 3
3 34( ) ( ) 4( )
4 4 .
y x x y x xy x x y x x
The result is the same as the original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.
Section 2.7 Graphing Techniques 231
Copyright © 2017 Pearson Education, Inc.
50. 3y x x Replace x with –x to obtain
3 3( ) ( ) .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to
obtain 3 3 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to
obtain 3 3( ) ( )y x x y x x 3 .y x x The result is the same as the
original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.
51. 2 8y x x Replace x with –x to obtain
2 2( ) ( ) 8 8.y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y
to obtain 2( ) ( ) 8y x x 2 28 8.y x x y x x
The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
52. y = x + 15 Replace x with –x to obtain
( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain ( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
53. 3 2f x x x
3
3 3
2
2 2
f x x xx x x x f x
The function is odd.
54. 5 32f x x x
5 3
5 3 5 3
2
2 2
f x x xx x x x f x
The function is odd.
55. 4 20.5 2 6f x x x
4 2
4 2
0.5 2 6
0.5 2 6
f x x xx x f x
The function is even.
56. 20.75 4f x x x
2
2
0.75 4
0.75 4
f x x xx x f x
The function is even.
57. 3 9f x x x
3
3 3
9
9 9
f x x xx x x x f x
The function is neither.
58. 4 5 8f x x x
4
4
5 8
5 8
f x x xx x f x
The function is neither.
59. 2 1f x x
This graph may be obtained by translating the
graph of 2y x 1 unit downward.
60. 2 2f x x
This graph may be obtained by translating the
graph of 2y x 2 units downward.
232 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
61. 2 2f x x
This graph may be obtained by translating the
graph of 2y x 2 units upward.
62. 2 3f x x
This graph may be obtained by translating the
graph of 2y x 3 units upward.
63. 24g x x
This graph may be obtained by translating the
graph of 2y x 4 units to the right.
64. 22g x x
This graph may be obtained by translating the
graph of 2y x 2 units to the right.
65. 22g x x
This graph may be obtained by translating the
graph of 2y x 2 units to the left.
66. 23g x x
This graph may be obtained by translating the
graph of 2y x 3 units to the left.
67. 1g x x
The graph is obtained by translating the graph of y x 1 unit downward.
68. 3 2g x x
This graph may be obtained by translating the graph of y x 3 units to the left and 2 units
upward.
Section 2.7 Graphing Techniques 233
Copyright © 2017 Pearson Education, Inc.
69. 3( 1)h x x
This graph may be obtained by translating the
graph of 3y x 1 unit to the left. It is then reflected across the x-axis.
70. 3( 1)h x x
This graph can be obtained by translating the
graph of 3y x 1 unit to the right. It is then reflected across the x-axis. (We may also reflect the graph about the x-axis first and then translate it 1 unit to the right.)
71. 22 1h x x
This graph may be obtained by translating the
graph of 2y x 1 unit down. It is then stretched vertically by a factor of 2.
72. 23 2h x x
This graph may be obtained by stretching the
graph of 2y x vertically by a factor of 3, then shifting the resulting graph down 2 units.
73. 22( 2) 4f x x
This graph may be obtained by translating the
graph of 2y x 2 units to the right and 4 units down. It is then stretched vertically by a factor of 2.
74. 23( 2) 1f x x
This graph may be obtained by translating the
graph of 2y x 2 units to the right and 1 unit up. It is then stretched vertically by a factor of 3 and reflected over the x-axis.
75. 2f x x
This graph may be obtained by translating the
graph of y x two units to the left.
234 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
76. 3f x x
This graph may be obtained by translating the
graph of y x three units to the right.
77. f x x
This graph may be obtained by reflecting the
graph of y x across the x-axis.
78. 2f x x
This graph may be obtained by translating the
graph of y x two units down.
79. 2 1f x x
This graph may be obtained by stretching the
graph of y x vertically by a factor of two and then translating the resulting graph one unit up.
80. 3 2f x x
This graph may be obtained by stretching the
graph of y x vertically by a factor of three and then translating the resulting graph two units down.
81. 312
4g x x
This graph may be obtained by stretching the
graph of 3y x vertically by a factor of 12
,
then shifting the resulting graph down four units.
82. 312
2g x x
This graph may be obtained by stretching the
graph of 3y x vertically by a factor of 12
,
then shifting the resulting graph up two units.
Section 2.7 Graphing Techniques 235
Copyright © 2017 Pearson Education, Inc.
83. 33g x x
This graph may be obtained by shifting the
graph of 3y x three units left.
84. 32f x x
This graph may be obtained by shifting the
graph of 3y x two units right.
85. 223
2f x x
This graph may be obtained by translating the
graph of 2y x two units to the right, then stretching the resulting graph vertically by a
factor of 23
.
86. Because ( ) ( ),g x x x f x the graphs
are the same.
87. (a) y = g(–x) The graph of g(x) is reflected across the y-axis.
(b) y = g(x – 2) The graph of g(x) is translated to the right 2 units.
(c) y = –g(x) The graph of g(x) is reflected across the x-axis.
(d) y = –g(x) + 2 The graph of g(x) is reflected across the x-axis and translated 2 units up.
236 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
88. (a) y f x
The graph of f(x) is reflected across the x-axis.
(b) 2y f x
The graph of f(x) is stretched vertically by a factor of 2.
(c) y f x
The graph of f(x) is reflected across the y-axis.
(d) 12
y f x
The graph of f(x) is compressed vertically
by a factor of 12
.
89. It is the graph of f x x translated 1 unit to
the left, reflected across the x-axis, and translated 3 units up. The equation is
1 3.y x
90. It is the graph of g x x translated 4 units
to the left, reflected across the x-axis, and translated two units up. The equation is
4 2.y x
91. It is the graph of f x x translated one
unit right and then three units down. The
equation is 1 3.y x
92. It is the graph of f x x translated 2 units
to the right, shrunken vertically by a factor of 12
, and translated one unit down. The
equation is 12
2 1.y x
93. It is the graph of g x x translated 4 units
to the left, stretched vertically by a factor of 2, and translated four units down. The equation
is 2 4 4.y x
94. It is the graph of f x x reflected across
the x-axis and then shifted two units down. The equation is 2y x .
95. Because f(3) = 6, the point (3, 6) is on the graph. Because the graph is symmetric with respect to the origin, the point (–3, –6) is on the graph. Therefore, f(–3) = –6.
96. Because f(3) = 6, (3, 6) is a point on the graph. The graph is symmetric with respect to the y-axis, so (–3, 6) is on the graph. Therefore, f(–3) = 6.
97. Because f(3) = 6, the point (3, 6) is on the graph. The graph is symmetric with respect to the line x = 6 and the point (3, 6) is 3 units to the left of the line x = 6, so the image point of (3, 6), 3 units to the right of the line x = 6 is (9, 6). Therefore, f(9) = 6.
98. Because f(3) = 6 and f(–x) = f(x), f(–3) = f(3). Therefore, f(–3) = 6.
99. Because (3, 6) is on the graph, (–3, –6) must also be on the graph. Therefore, f(–3) = –6.
100. If f is an odd function, f(–x) = –f(x). Because f(3) = 6 and f(–x) = –f(x), f(–3) = –f(3). Therefore, f(–3) = –6.
Chapter 2 Quiz (Sections 2.5−2.7) 237
Copyright © 2017 Pearson Education, Inc.
101. f(x) = 2x + 5 Translate the graph of ( )f x up 2 units to obtain the graph of
( ) (2 5) 2 2 7.t x x x Now translate the graph of t(x) = 2x + 7 left 3 units to obtain the graph of
( ) 2( 3) 7 2 6 7 2 13.g x x x x (Note that if the original graph is first translated to the left 3 units and then up 2 units, the final result will be the same.)
102. f(x) = 3 – x Translate the graph of ( )f x down 2 units to
obtain the graph of ( ) (3 ) 2 1.t x x x
Now translate the graph of 1t x x right
3 units to obtain the graph of ( ) ( 3) 1 3 1 4.g x x x x
(Note that if the original graph is first translated to the right 3 units and then down 2 units, the final result will be the same.)
103. (a) Because f(–x) = f(x), the graph is symmetric with respect to the y-axis.
(b) Because f(–x) = –f(x), the graph is symmetric with respect to the origin.
104. (a) f(x) is odd. An odd function has a graph symmetric with respect to the origin. Reflect the left half of the graph in the origin.
(b) f(x) is even. An even function has a graph symmetric with respect to the y-axis. Reflect the left half of the graph in the y-axis.
Chapter 2 Quiz (Sections 2.5−2.7)
1. (a) First, find the slope: 9 5
21 ( 3)
m
Choose either point, say, (−3, 5), to find the equation of the line:
5 2( ( 3)) 2( 3) 5y x y x 2 11y x .
(b) To find the x-intercept, let y = 0 and solve
for x: 112
0 2 11x x . The
x-intercept is 112
, 0 .
2. Write 3x − 2y = 6 in slope-intercept form to
find its slope: 32
3 2 6 3.x y y x
Then, the slope of the line perpendicular to
this graph is 23
. 23
4 ( ( 6))y x
2 23 3
( 6)) 4y x y x
3. (a) 8x (b) 5y
4. (a) Cubing function; domain: ( , ) ;
range: ( , ) ; increasing over
( , ) .
(b) Absolute value function; domain: ( , ) ; range: [0, ) ; decreasing over
( , 0) ; increasing over (0, )
238 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(c) Cube root function: domain: ( , ) ;
range: ( , ) ; increasing over
( , ) .
5.
0.40 0.75
5.5 0.40 5.5 0.75
0.40 5 0.75 2.75
f x xf
A 5.5-minute call costs $2.75.
6. if 0( )2 3 if 0
x xf xx x
For values of x < 0, the graph is the line y = 2x + 3. Do not include the right endpoint
(0, 3). Graph the line y x for values of x ≥ 0, including the left endpoint (0, 0).
7. 3( ) 1f x x
Reflect the graph of 3( )f x x across the x-axis, and then translate the resulting graph one unit up.
8. ( ) 2 1 3f x x
Shift the graph of ( )f x x one unit right,
stretch the resulting graph vertically by a factor of 2, then shift this graph three units up.
9. This is the graph of ( )g x x , translated four units to the left, reflected across the x-axis, and then translated two units down.
The equation is 4 2y x .
10. (a) 2 7f x x
Replace x with –x to obtain
2
2
( ) 7
7
f x xf x x f x
The result is the same as the original function, so the function is even.
(b) 3 1f x x x
Replace x with –x to obtain
3
3
1
1
f x x xx x f x
The result is not the same as the original equation, so the function is not even. Because ,f x f x the function is
not odd. Therefore, the function is neither even nor odd.
(c) 101 99f x x x
Replace x with –x to obtain
101 99
101 99
101 99
f x x xx x
x xf x
Because f(−x) = −f(x), the function is odd.
Section 2.8 Function Operations and Composition
In exercises 1–10, 1f x x and 2.g x x
1. 2
2 2 2
2 1 2 7
f g f g
2. 2
2 2 2
2 1 2 1
f g f g
3. 2
2 2 2
2 1 2 12
fg f g
4. 2
2 2 1 32
2 42
ffg g
5. 2 22 2 2 2 1 5f g f g f
Section 2.8 Function Operations and Composition 239
Copyright © 2017 Pearson Education, Inc.
6. 22 2 2 1 2 1 9g f g f g
7. f is defined for all real numbers, so its domain is , .
8. g is defined for all real numbers, so its domain is , .
9. f + g is defined for all real numbers, so its domain is , .
10. fg
is defined for all real numbers except those
values that make 0,g x so its domain is
, 0 0, .
In Exercises 11–18, 2( ) 3f x x and
( ) 2 6g x x .
11.
2
( )(3) (3) (3)
(3) 3 2(3) 6
12 0 12
f g f g
12. 2
( )( 5) ( 5) ( 5)
[( 5) 3] [ 2( 5) 6]28 16 44
f g f g
13. 2
( )( 1) ( 1) ( 1)
[( 1) 3] [ 2( 1) 6]4 8 4
f g f g
14. 2
( )(4) (4) (4)
[(4) 3] [ 2(4) 6]19 ( 2) 21
f g f g
15. 2
( )(4) (4) (4)
[4 3] [ 2(4) 6]19 ( 2) 38
fg f g
16. 2
( )( 3) ( 3) ( 3)
[( 3) 3] [ 2( 3) 6]12 12 144
fg f g
17. 2( 1) ( 1) 3 4 1
( 1)( 1) 2( 1) 6 8 2
f fg g
18. 2(5) (5) 3 28
(5) 7(5) 2(5) 6 4
f fg g
19. ( ) 3 4, ( ) 2 5f x x g x x
( )( ) ( ) ( )(3 4) (2 5) 5 1
f g x f x g xx x x
( )( ) ( ) ( )(3 4) (2 5) 9
f g x f x g xx x x
2
2
( )( ) ( ) ( ) (3 4)(2 5)
6 15 8 20
6 7 20
fg x f x g x x xx x xx x
( ) 3 4( )
( ) 2 5
f f x xxg g x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g,
and fg are all , . The domain of fg is
the set of all real numbers for which 0.g x This is the set of all real numbers
except 52
, which is written in interval notation
as 5 52 2
, , .
20. f(x) = 6 – 3x, g(x) = –4x + 1 ( )( ) ( ) ( )
(6 3 ) ( 4 1)7 7
f g x f x g xx x
x
( )( ) ( ) ( )(6 3 ) ( 4 1) 5
f g x f x g xx x x
2
2
( )( ) ( ) ( ) (6 3 )( 4 1)
24 6 12 3
12 27 6
fg x f x g x x xx x x
x x
( ) 6 3( )
( ) 4 1
f f x xxg g x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all , . The domain of
fg is the set of all real numbers for which
0.g x This is the set of all real numbers
except 14
, which is written in interval notation
as 1 14 4
– , , .
21. 2 2( ) 2 3 , ( ) 3f x x x g x x x
2 2
2
( )( ) ( ) ( )
(2 3 ) ( 3)
3 4 3
f g x f x g xx x x x
x x
2 2
2 2
2
( )( ) ( ) ( )
(2 3 ) ( 3)
2 3 3
2 3
f g x f x g xx x x x
x x x xx x
(continued on next page)
240 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
2 2
4 3 2 3 2
4 3 2
( )( ) ( ) ( )
(2 3 )( 3)
2 2 6 3 3 9
2 5 9 9
fg x f x g xx x x x
x x x x x xx x x x
2
2
( ) 2 3( )
( ) 3
f f x x xxg g x x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all , . The domain of
fg is the set of all real numbers for which
0.g x If 2 3 0x x , then by the
quadratic formula 1 112ix . The equation
has no real solutions. There are no real numbers which make the denominator zero.
Thus, the domain of fg is also , .
22. 2 2( ) 4 2 , ( ) 3 2f x x x g x x x
2 2
2
( )( ) ( ) ( )
(4 2 ) ( 3 2)
5 2
f g x f x g xx x x x
x x
2 2
2 2
2
( )( ) ( ) ( )
(4 2 ) ( 3 2)
4 2 3 2
3 5 2
f g x f x g xx x x x
x x x xx x
2 2
4 3 2 3 2
4 3 2
( )( ) ( ) ( )
(4 2 )( 3 2)
4 12 8 2 6 4
4 10 2 4
fg x f x g xx x x x
x x x x x xx x x x
2
2
( ) 4 2( )
( ) 3 2
f f x x xxg g x x x
The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g,
and fg are all , . The domain of fg is the
set of all real numbers x such that 2 3 2 0x x . Because 2 3 2 ( 1)( 2)x x x x , the numbers
which give this denominator a value of 0 are
x = 1 and x = 2. Therefore, the domain of fg is
the set of all real numbers except 1 and 2, which is written in interval notation as (– , 1) (1, 2) (2, ) .
23. 1
( ) 4 1, ( )f x x g xx
1( )( ) ( ) ( ) 4 1f g x f x g x x
x
1( )( ) ( ) ( ) 4 1f g x f x g x x
x
( )( ) ( ) ( )
1 4 14 1
fg x f x g xxx
x x
1
( ) 4 1( ) 4 1
( ) x
f f x xx x xg g x
Because 14
4 1 0 4 1 ,x x x the
domain of f is 14
, . The domain of g is
, 0 0, . Considering the intersection
of the domains of f and g, the domains of f + g,
f – g, and fg are all 14
, . Because 1 0x
for any value of x, the domain of fg is also
14
, .
24. 1
( ) 5 4, ( )f x x g xx
( )( ) ( ) ( )1 1
5 4 5 4
f g x f x g x
x xx x
( )( ) ( ) ( )1 1
5 4 5 4
f g x f x g x
x xx x
( )( ) ( ) ( )
1 5 45 4
fg x f x g xxx
x x
1
( ) 5 4( ) 5 4
( ) x
f f x xx x xg g x
Because 45
5 4 0 5 4 ,x x x the
domain of f is 45
, . The domain of g is
, 0 0, . Considering the intersection
of the domains of f and g, the domains of f + g,
f – g, and fg are all 45
, . 1 0x for any
value of x, so the domain of fg is also
45
, .
Section 2.8 Function Operations and Composition 241
Copyright © 2017 Pearson Education, Inc.
25. 2008 280 and 2008 470, thus
2008 2008 2008280 470 750 (thousand).
M FT M F
26. 2012 390 and 2012 630, thus
2012 2012 2012390 630 1020 (thousand).
M FT M F
27. Looking at the graphs of the functions, the slopes of the line segments for the period 2008−2012 are much steeper than the slopes of the corresponding line segments for the period 2004−2008. Thus, the number of associate’s degrees increased more rapidly during the period 2008−2012.
28. If 2004 2012, ( ) ,k T k r and F(k) = s, then M(k) = r − s.
29. 2000 2000 200019 13 6
T S T S
It represents the dollars in billions spent for general science in 2000.
30. 2010 2010 201029 11 18
T G T G
It represents the dollars in billions spent on space and other technologies in 2010.
31. Spending for space and other technologies spending decreased in the years 1995−2000 and 2010–2015.
32. Total spending increased the most during the years 2005−2010.
33. (a)
2 2 2
4 2 2
f g f g
(b) ( )(1) (1) (1) 1 ( 3) 4f g f g
(c) ( )(0) (0) (0) 0( 4) 0fg f g
(d) (1) 1 1
(1)(1) 3 3
f fg g
34. (a) ( )(0) (0) (0) 0 2 2f g f g
(b) ( )( 1) ( 1) ( 1)2 1 3
f g f g
(c) ( )(1) (1) (1) 2 1 2fg f g
(d) (2) 4
(2) 2(2) 2
f fg g
35. (a) ( )( 1) ( 1) ( 1) 0 3 3f g f g
(b) ( )( 2) ( 2) ( 2)1 4 5
f g f g
(c) ( )(0) (0) (0) 1 2 2fg f g
(d) (2) 3
(2) undefined(2) 0
f fg g
36. (a) ( )(1) (1) (1) 3 1 2f g f g
(b) ( )(0) (0) (0) 2 0 2f g f g
(c) ( )( 1) ( 1) ( 1) 3( 1) 3fg f g
(d) (1) 3
(1) 3(1) 1
f fg g
37. (a) 2 2 2 7 2 5f g f g
(b) ( )(4) (4) (4) 10 5 5f g f g
(c) ( )( 2) ( 2) ( 2) 0 6 0fg f g
(d) (0) 5
(0) undefined(0) 0
f fg g
38. (a) 2 2 2 5 4 9f g f g
(b) ( )(4) (4) (4) 0 0 0f g f g
(c) ( )( 2) ( 2) ( 2) 4 2 8fg f g
(d) (0) 8
(0) 8(0) 1
f fg g
242 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
39. x f x g x f g x f g x fg x f xg
−2 0 6 0 6 6 0 6 6 0 6 0 06
0
0 5 0 5 0 5 5 0 5 5 0 0 50
undefined
2 7 −2 7 2 5 7 2 9 7 2 14 72
3.5
4 10 5 10 5 15 10 5 5 10 5 50 105
2
40. x f x g x f g x f g x fg x f xg
−2 −4 2 4 2 2 4 2 6 4 2 8 42
2
0 8 −1 8 1 7 8 1 9 8 1 8 81
8
2 5 4 5 4 9 5 4 1 5 4 20 54
1.25
4 0 0 0 0 0 0 0 0 0 0 0 00
undefined
41. Answers may vary. Sample answer: Both the
slope formula and the difference quotient represent the ratio of the vertical change to the horizontal change. The slope formula is stated for a line while the difference quotient is stated for a function f.
42. Answers may vary. Sample answer: As h approaches 0, the slope of the secant line PQ approaches the slope of the line tangent of the curve at P.
43. 2f x x
(a) ( ) 2 ( ) 2f x h x h x h
(b) ( ) ( ) (2 ) (2 )2 2
f x h f x x h xx h x h
(c) ( ) ( )
1f x h f x h
h h
44. 1f x x
(a) ( ) 1 ( ) 1f x h x h x h
(b) ( ) ( ) (1 ) (1 )1 1
f x h f x x h xx h x h
(c) ( ) ( )
1f x h f x h
h h
45. 6 2f x x
(a) ( ) 6( ) 2 6 6 2f x h x h x h
(b) ( ) ( )(6 6 2) (6 2)6 6 2 6 2 6
f x h f xx h x
x h x h
(c) ( ) ( ) 6
6f x h f x h
h h
46. 4 11f x x
(a) ( ) 4( ) 11 4 4 11f x h x h x h
(b) ( ) ( )(4 4 11) (4 11)4 4 11 4 11 4
f x h f xx h x
x h x h
(c) ( ) ( ) 4
4f x h f x h
h h
47. 2 5f x x
(a) ( ) 2( ) 52 2 5
f x h x hx h
(b) ( ) ( )( 2 2 5) ( 2 5)
2 2 5 2 5 2
f x h f xx h x
x h x h
(c) ( ) ( ) 2
2f x h f x h
h h
Section 2.8 Function Operations and Composition 243
Copyright © 2017 Pearson Education, Inc.
48. ( ) 4 2f x x
(a) ( ) 4( ) 24 4 2
f x h x hx h
(b)
( ) ( )4 4 2 4 24 4 2 4 24
f x h f xx h xx h xh
(c) ( ) ( ) 4
4f x h f x h
h h
49. 1
( )f xx
(a) 1
( )f x hx h
(b)
( ) ( )
1 1
f x h f xx x h
x h x x x hh
x x h
(c)
( ) ( )
1
hx x hf x h f x h
h h hx x h
x x h
50. 2
1( )f x
x
(a) 2
1( )f x h
x h
(b)
22
2 2 22
2 2 2 2
2 22 2
( ) ( )
1 1
2 2
f x h f xx x h
xx h x x hx x xh h xh h
x x h x x h
(c)
2
22
22
22
22
22
( ) ( ) 2
2
2
xh hx x hf x h f x xh h
h h hx x hh x h
hx x hx h
x x h
51. 2( )f x x
(a) 2 2 2( ) ( ) 2f x h x h x xh h
(b) 2 2 2
2( ) ( ) 2
2
f x h f x x xh h xxh h
(c) 2( ) ( ) 2
(2 )
2
f x h f x xh hh h
h x hh
x h
52. 2( )f x x
(a)
2
2 2
2 2
( ) ( )
2
2
f x h x hx xh h
x xh h
(b) 2 2 2
2 2 2
2
( ) ( ) 2
2
2
f x h f x x xh h x
x xh h xxh h
(c) 2( ) ( ) 2
(2 )
2
f x h f x xh hh h
h x hh
x h
53. 2( ) 1f x x
(a) 2
2 2
2 2
( ) 1 ( )
1 ( 2 )
1 2
f x h x hx xh h
x xh h
(b) 2 2 2
2 2 2
2
( ) ( )
(1 2 ) (1 )
1 2 1
2
f x h f xx xh h x
x xh h xxh h
(c) 2( ) ( ) 2
( 2 )
2
f x h f x xh hh h
h x hh
x h
54. 2( ) 1 2f x x
(a) 2
2 2
2 2
( ) 1 2( )
1 2( 2 )
1 2 4 2
f x h x hx xh h
x xh h
244 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b)
2 2 2
2 2 2
2
( ) ( )
1 2 4 2 1 2
1 2 4 2 1 2
4 2
f x h f xx xh h x
x xh h xxh h
(c) 2( ) ( ) 4 2
(4 2 )
4 2
f x h f x xh hh h
h x hh
x h
55. 2( ) 3 1f x x x
(a) 2
2 2
( ) 3 1
2 3 3 1
f x h x h x hx xh h x h
(b)
2 2
2
2 2 2
2
( ) ( )
2 3 3 1
3 1
2 3 3 1 3 1
2 3
f x h f xx xh h x h
x x
x xh h x h x xxh h h
(c) 2( ) ( ) 2 3
(2 3)
2 3
f x h f x xh h hh h
h x hh
x h
56. 2( ) 4 2f x x x
(a) 2
2 2
( ) 4 2
2 4 4 2
f x h x h x hx xh h x h
(b)
2 2
2
2 2 2
2
( ) ( )
2 4 4 2
4 2
2 4 4 2 4 2
2 4
f x h f xx xh h x h
x x
x xh h x h x xxh h h
(c) 2( ) ( ) 2 4
(2 4)
2 4
f x h f x xh h hh h
h x hh
x h
57. 3 4 4 3 1g x x g
4 4 1
2 1 3 2 3 5
f g f g f
58. 3 2 2 3 1g x x g
2 2 1
2 1 3 2 3 1
f g f g f
59. 3 2 2 3 5g x x g
2 2 5
2 5 3 10 3 7
f g f g f
60. 2 3f x x 3 2 3 3 6 3 3f
3 3 3 3 3 0g f g f g
61. 2 3 0 2 0 3 0 3 3f x x f
0 0 3
3 3 3 3 6
g f g f g
62. 2 3 2 2 2 3 7f x x f
2 2 7
7 3 7 3 10
g f g f g
63. 2 3 2 2 2 3 4 3 1f x x f
2 2 1 2 1 3 1f f f f f
64. 3 2 2 3 5g x x g
2 2 5 5 3 2g g g g g
65. ( )(2) [ (2)] (3) 1f g f g f
66. ( )(7) [ (7)] (6) 9f g f g f
67. ( )(3) [ (3)] (1) 9g f g f g
68. ( )(6) [ (6)] (9) 12g f g f g
69. ( )(4) [ (4)] (3) 1f f f f f
70. ( )(1) [ (1)] (9) 12g g g g g
71. ( )(1) [ (1)] (9)f g f g f However, f(9) cannot be determined from the table given.
72. ( ( ))(7) ( ( (7)))( (6)) (9) 12
g f g g f gg f g
73. (a) ( )( ) ( ( )) (5 7)6(5 7) 930 42 9 30 33
f g x f g x f xx
x x
The domain and range of both f and g are ( , ) , so the domain of f g is
( , ) .
Section 2.8 Function Operations and Composition 245
Copyright © 2017 Pearson Education, Inc.
(b) ( )( ) ( ( )) ( 6 9)5( 6 9) 7
30 45 7 30 52
g f x g f x g xx
x x
The domain of g f is ( , ) .
74. (a) ( )( ) ( ( )) (3 1)8(3 1) 1224 8 12 24 4
f g x f g x f xxx x
The domain and range of both f and g are ( , ) , so the domain of f g is
( , ) .
(b) ( )( ) ( ( )) (8 12)3(8 12) 124 36 1 24 35
g f x g f x g xxx x
The domain of g f is ( , ) .
75. (a) ( )( ) ( ( )) ( 3) 3f g x f g x f x x
The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 3 0 3x x .
Therefore, the domain of f g is
[ 3, ) .
(b) ( )( ) ( ( )) 3g f x g f x g x x
The domain and range of g are ( , ) , however, the domain and range of f are [0, ) .Therefore, the domain of g f is
[0, ) .
76. (a) ( )( ) ( ( )) ( 1) 1f g x f g x f x x
The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 1 0 1x x . Therefore,
the domain of f g is [1, ) .
(b) ( )( ) ( ( )) 1g f x g f x g x x
The domain and range of g are ( , ), however, the domain and range of f are [0, ). Therefore, the domain of g f is
[0, ).
77. (a) 2
2 3
( )( ) ( ( )) ( 3 1)
( 3 1)
f g x f g x f x xx x
The domain and range of f and g are ( , ), so the domain of f g is
( , ).
(b)
3
23 3
6 3
( )( ) ( ( )) ( )
3 1
3 1
g f x g f x g x
x x
x x
The domain and range of f and g are ( , ), so the domain of g f is
( , ).
78. (a) 4 2
4 2
4 2
( )( ) ( ( )) ( 4)
4 2
2
f g x f g x f x xx xx x
The domain of f and g is ( , ), while
the range of f is ( , ) and the range of
g is 4, , so the domain of f g is
( , ).
(b) 4 2
( )( ) ( ( )) ( 2)
( 2) ( 2) 4
g f x g f x g xx x
The domain of f and g is ( , ), while
the range of f is ( , ) and the range of
g is 4, , so the domain of g f is
( , ).
79. (a) ( )( ) ( ( )) (3 ) 3 1f g x f g x f x x
The domain and range of g are ( , ) ,
however, the domain of f is [1, ), while
the range of f is [0, ). So, 13
3 1 0x x . Therefore, the
domain of f g is 13
, .
(b) ( )( ) ( ( )) 1
3 1
g f x g f x g x
x
The domain and range of g are ( , ) ,
however, the range of f is [0, ) . So
1 0 1x x . Therefore, the domain of g f is [1, ) .
80. (a) ( )( ) ( ( )) (2 ) 2 2f g x f g x f x x
The domain and range of g are ( , ) ,
however, the domain of f is [2, ) . So,
2 2 0 1x x . Therefore, the
domain of f g is 1, .
246 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(b) ( )( ) ( ( )) 2g f x g f x g x
2 2x
The domain and range of g are ( , ) ,
however, the range of f is [0, ) . So
2 0 2x x . Therefore, the domain of g f is [2, ) .
81. (a) 21
( )( ) ( ( )) ( 1) xf g x f g x f x
The domain and range of g are ( , ) , however, the domain of f is ( ,0) (0, ) . So, 1 0 1x x .
Therefore, the domain of f g is
( , 1) ( 1, ) .
(b) 2 2( )( ) ( ( )) 1x xg f x g f x g
The domain and range of f is ( ,0) (0, ) , however, the domain
and range of g are ( , ) . So 0x .
Therefore, the domain of g f is
( ,0) (0, ) .
82. (a) 44
( )( ) ( ( )) ( 4) xf g x f g x f x
The domain and range of g are ( , ) , however, the domain of f is ( ,0) (0, ) . So, 4 0 4x x .
Therefore, the domain of f g is
( , 4) ( 4, ) .
(b) 4 4( )( ) ( ( )) 4x xg f x g f x g
The domain and range of f is ( ,0) (0, ) , however, the domain
and range of g are ( , ) . So 0x .
Therefore, the domain of g f is
( ,0) (0, ) .
83. (a) 1 1( )( ) ( ( )) 2x xf g x f g x f
The domain and range of g are ( , 0) (0, ) , however, the domain
of f is [ 2, ) . So, 1 2 0x
12
0 or x x (using test intervals).
Therefore, the domain of f g is
12
, 0 , .
(b) 12
( )( ) ( ( )) 2x
g f x g f x g x
The domain of f is [ 2, ) and its range is
[0, ). The domain and range of g
are ( ,0) (0, ) . So 2 0 2x x .
Therefore, the domain of g f is ( 2, ) .
84. (a) 2 2( )( ) ( ( )) 4x xf g x f g x f
The domain and range of g are ( , 0) (0, ) , however, the domain
of f is [ 4, ) . So, 2 4 0x
12
0 or x x (using test intervals).
Therefore, the domain of f g is
12
, 0 , .
(b) 24
( )( ) ( ( )) 4x
g f x g f x g x
The domain of f is [ 4, ) and its range is
[0, ). The domain and range of g
are ( , 0) (0, ) . So 4 0 4x x .
Therefore, the domain of g f is ( 4, ) .
85. (a) 1 15 5
( )( ) ( ( )) x xf g x f g x f
The domain of g is ( , 5) ( 5, ) ,
and the range of g is ( , 0) (0, ) .
The domain of f is [0, ) . Therefore, the
domain of f g is 5, .
(b) 15
( )( ) ( ( ))x
g f x g f x g x
The domain and range of f is [0, ). The
domain of g is ( , 5) ( 5, ).
Therefore, the domain of g f is [0, ).
86. (a) 3 36 6
( )( ) ( ( )) x xf g x f g x f
The domain of g is ( , 6) ( 6, ) ,
and the range of g is ( , 0) (0, ) .
The domain of f is [0, ) . Therefore, the
domain of f g is 6, .
(b) 36
( )( ) ( ( ))x
g f x g f x g x
The domain and range of f is [0, ) . The
domain of g is ( , 6) ( 6, ) .
Therefore, the domain of g f is [0, ).
Section 2.8 Function Operations and Composition 247
Copyright © 2017 Pearson Education, Inc.
87. (a) 1 11 2 1 2
( )( ) ( ( )) xx x xf g x f g x f
The domain and range of g are ( , 0) (0, ) . The domain of f is
( , 2) ( 2, ), and the range of f is
( , 0) (0, ) . So, 1 2
0 0xx x or
12
0 x or 12
x (using test intervals).
Thus, 0x and 12
x . Therefore, the
domain of f g is
1 12 2
, 0 0, , .
(b) 1 12 1 ( 2)
( )( ) ( ( ))
2x xg f x g f x g
x
The domain and range of g are ( , 0) (0, ). The domain of f is
( , 2) (2, ), and the range of f is
( , 0) (0, ). Therefore, the domain of
g f is ( , 2) (2, ).
88. (a) 1 11 4
1 4
( )( ) ( ( )) x xx
x
f g x f g x f
The domain and range of g are ( , 0) (0, ). The domain of f is
( , 4) ( 4, ), and the range of f is
( , 0) (0, ). So, 1 4
0 0xx x
or 14
0 x or 14
1 4 0x x
(using test intervals). Thus, x ≠ 0 and
x ≠ 14
. Therefore, the domain of f g is
1 14 4
, 0 0, , .
(b) 1 14 1 ( 4)
( )( ) ( ( ))
4x xg f x g f x g
x
The domain and range of g are ( ,0) (0, ) . The domain of f is
( , 4) ( 4, ) , and the range of f is
( ,0) (0, ) . Therefore, the domain of
g f is ( , 4) ( 4, ).
89.
(2) (1) 2 and (3) (2) 5
Since (1) 7 and (1) 3, (3) 7.
g f g g f gg f f g
x f x g x ( )g f x
1 3 2 7
2 1 5 2
3 2 7 5
90. ( )f x is odd, so ( 1) (1) ( 2) 2.f f
Because ( )g x is even, (1) ( 1) 2g g and
(2) ( 2) 0.g g ( )( 1) 1,f g so
( 1) 1f g and (2) 1.f ( )f x is odd, so
( 2) (2) 1.f f Thus,
( )( 2) ( 2) (0) 0f g f g f and
( )(1) (1) (2) 1f g f g f and
( )(2) (2) (0) 0.f g f g f
x –2 –1 0 1 2
f x –1 2 0 –2 1
g x 0 2 1 2 0
f g x 0 1 –2 1 0
91. Answers will vary. In general, composition of functions is not commutative. Sample answer:
2 3 3 2 3 26 9 2 6 11
3 2 2 3 2 36 4 3 6 7
f g x f x xx x
g f x g x xx x
Thus, f g x is not equivalent to
g f x .
92.
3
33
7
7 7
7 7
f g x f g x f x
xx x
3
33 33
7
7 7
g f x g f x g x
x x x
93.
14
14
4 2 2
4 2 2
2 2 2 2
f g x f g x x
xx x x
14
1 14 4
4 2 2
4 2 2 4
g f x g f x xx x x
94.
13
13
3
3
f g x f g x x
x x
13
13
3
3
g f x g f x x
x x
248 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
95. 31 435 5
3 33 3
5 4
4 4
f g x f g x x
x x x
31 435 5
51 4 4 45 5 5 5 555
5 4
5 4 x
x
g f x g f x x
xx
96. 33
3 33 3
1 1
1 1
f g x f g x x
x x x
33 1 1
1 1
g f x g f x xx x
In Exercises 97−102, we give only one of many possible answers.
97. 2( ) (6 2)h x x
Let g(x) = 6x – 2 and 2( )f x x . 2( )( ) (6 2) (6 2) ( )f g x f x x h x
98. 2 2( ) (11 12 )h x x x 2 2Let ( ) 11 12 and ( ) .g x x x f x x
2
2 2
( )( ) (11 12 )
(11 12 ) ( )
f g x f x xx x h x
99. 2( ) 1h x x
Let 2( ) 1 and ( ) .g x x f x x
2 2( )( ) ( 1) 1 ( ).f g x f x x h x
100. 3( ) (2 3)h x x 3Let ( ) 2 3 and ( ) .g x x f x x
3( )( ) (2 3) (2 3) ( )f g x f x x h x
101. ( ) 6 12h x x
Let ( ) 6 and ( ) 12.g x x f x x
( )( ) (6 ) 6 12 ( )f g x f x x h x
102. 3( ) 2 3 4h x x
Let 3( ) 2 3 and ( ) 4.g x x f x x 3( )( ) (2 3) 2 3 4 ( )f g x f x x h x
103. f(x) = 12x, g(x) = 5280x ( )( ) [ ( )] (5280 )
12(5280 ) 63,360f g x f g x f x
x x
The function f g computes the number of inches in x miles.
104. 3 , 1760f x x g x x
1760
3 1760 5280
f g x f g x f xx x
f g x compute the number of feet in x
miles.
105. 23( )
4x x
(a) 2 2 23 3(2 ) (2 ) (4 ) 3
4 4x x x x
(b) 2(16) (2 8) 3(8)
64 3 square units
A
106. (a) 4 4
4 x xx s s s
(b) 2224 16x xy s
(c) 26 36
2.25 square units16 16
y
107. (a) 2( ) 4 and ( )r t t r r
2 2( )( ) [ ( )]
(4 ) (4 ) 16
r t r tt t t
(b) ( )( )r t defines the area of the leak in terms of the time t, in minutes.
(c) 2 2(3) 16 (3) 144 ft
108. (a) 2 2
( )( ) [ ( )]
(2 ) (2 ) 4
r t r tt t t
(b) It defines the area of the circular layer in terms of the time t, in hours.
(c) 2 2( )(4) 4 (4) 64 mir
109. Let x = the number of people less than 100 people that attend.
(a) x people fewer than 100 attend, so 100 – x people do attend N(x) = 100 – x
(b) The cost per person starts at $20 and increases by $5 for each of the x people that do not attend. The total increase is $5x, and the cost per person increases to $20 + $5x. Thus, G(x) = 20 + 5x.
(c) ( ) ( ) ( ) (100 )(20 5 )C x N x G x x x
Chapter 2 Review Exercises 249
Copyright © 2017 Pearson Education, Inc.
(d) If 80 people attend, x = 100 – 80 = 20.
20 100 20 20 5 20
80 20 100
80 120 $9600
C
110. (a) 1 0.02y x
(b) 2 0.015( 500)y x
(c) 1 2y y represents the total annual interest.
(d) 1 2
1 2
( )(250)(250) (250)
0.02(250) 0.015(250 500)5 0.015(750) 15 11.25$16.25
y yy y
111. (a) 1
2g x x
(b) 1f x x
(c) 1 11
2 2f g x f g x f x x
(d) 160 60 1 $31
2f g
112. If the area of a square is 2x square inches, each side must have a length of x inches. If 3 inches is added to one dimension and 1 inch is subtracted from the other, the new dimensions will be x + 3 and x – 1. Thus, the area of the resulting rectangle is (x) = (x + 3)(x – 1).
Chapter 2 Review Exercises
1. P(3, –1), Q(–4, 5) 2 2
2 2
( , ) ( 4 3) [5 ( 1)]
( 7) 6 49 36 85
d P Q
Midpoint:
3 ( 4) 1 5 1 4 1, , , 2
2 2 2 2 2
2. M(–8, 2), N(3, –7) 2 2
2 2
( , ) [3 ( 8)] ( 7 2)
11 ( 9) 121 81 202
d M N
Midpoint: 8 3 2 ( 7) 5 5
, , 2 2 2 2
3. A(–6, 3), B(–6, 8) 2 2
2
( , ) [ 6 ( 6)] (8 3)
0 5 25 5
d A B
Midpoint: 6 ( 6) 3 8 12 11 11
, , 6,2 2 2 2 2
4. Label the points A(5, 7), B(3, 9), and C(6, 8).
2 2
2 2
( , ) 3 5 9 7
2 2 4 4 8
d A B
2 2
2 2
( , ) (6 5) (8 7)
1 1 1 1 2
d A C
2 2
22
( , ) 6 3 8 9
3 1 9 1 10
d B C
Because 2 2 28 2 10 , triangle
ABC is a right triangle with right angle at (5, 7).
5. Let B have coordinates (x, y). Using the midpoint formula, we have
6 108, 2 ,
2 26 10
8 22 2
6 16 10 422 6
x y
x y
x yx y
The coordinates of B are (22, −6).
6. P(–2, –5), Q(1, 7), R(3, 15) 2 2
2 2
( , ) (1 ( 2)) (7 ( 5))
(3) (12) 9 144
153 3 17
d P Q
2 2
2 2
( , ) (3 1) (15 7)
2 8 4 64 68 2 17
d Q R
2 2
2 2
( , ) (3 ( 2)) (15 ( 5))
(5) (20) 25 400 5 17
d P R
(continued on next page)
250 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
(continued)
( , ) ( , ) 3 17 2 17d P Q d Q R
5 17 ( , ),d P R so these three points are collinear.
7. Center (–2, 3), radius 15 2 2 2
2 2 2
2 2
( ) ( )
[ ( 2)] ( 3) 15
( 2) ( 3) 225
x h y k rx y
x y
8. Center ( 5, 7 ), radius 3
2 2 2
22 2
2 2
( ) ( )
5 7 3
5 7 3
x h y k r
x y
x y
9. Center (–8, 1), passing through (0, 16) The radius is the distance from the center to any point on the circle. The distance between (–8, 1) and (0, 16) is
2 2 2 2(0 ( 8)) (16 1) 8 15
64 225 289 17.
r
The equation of the circle is 2 2 2
2 2[ ( 8)] ( 1) 17
( 8) ( 1) 289
x yx y
10. Center (3, –6), tangent to the x-axis The point (3, –6) is 6 units directly below the x-axis. Any segment joining a circle’s center to a point on the circle must be a radius, so in this case the length of the radius is 6 units.
2 2 2
2 2 2
2 2
( ) ( )
( 3) [ ( 6)] 6
( 3) ( 6) 36
x h y k rx y
x y
11. The center of the circle is (0, 0). Use the distance formula to find the radius:
2 2 2(3 0) (5 0) 9 25 34r
The equation is 2 2 34x y .
12. The center of the circle is (0, 0). Use the distance formula to find the radius:
2 2 2( 2 0) (3 0) 4 9 13r
The equation is 2 2 13x y .
13. The center of the circle is (0, 3). Use the distance formula to find the radius:
2 2 2( 2 0) (6 3) 4 9 13r
The equation is 2 2( 3) 13x y .
14. The center of the circle is (5, 6). Use the distance formula to find the radius:
2 2 2(4 5) (9 6) 1 9 10r
The equation is 2 2( 5) ( 6) 10x y .
15. 2 24 6 12 0x x y y Complete the square on x and y to put the equation in center-radius form.
2 2
2 2
2 2
4 6 12
4 4 6 9 12 4 9
2 3 1
x x y y
x x y y
x y
The circle has center (2, –3) and radius 1.
16. 2 26 10 30 0x x y y Complete the square on x and y to put the equation in center-radius form.
2 2
2 2
( 6 9) ( 10 25) 30 9 25
( 3) ( 5) 4
x x y yx y
The circle has center (3, 5) and radius 2.
17.
2 2
2 2
2 2
2 249 9 49 94 4 4 4
2 27 3 49 942 2 4 4 4
2 27 3 542 2 4
2 14 2 6 2 0
7 3 1 0
7 3 1
7 3 1
x x y yx x y yx x y y
x x y y
x y
x y
The circle has center 7 32 2
, and radius
54 9 6 3 6544 24 4
.
18.
2 2
2 2
2 2
2 2 25 25121 1214 4 4 4
22 5 146112 2 4
3 33 3 15 0
11 5 0
11 5 0
11 5 0
x x y yx x y y
x x y y
x x y y
x y
The circle has center 5112 2
, and radius
1462
.
19. This is not the graph of a function because a vertical line can intersect it in two points. domain: [−6, 6]; range: [−6, 6]
20. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: 0,
Chapter 2 Review Exercises 251
Copyright © 2017 Pearson Education, Inc.
21. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: ( , 1] [1, )
22. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: 0,
23. This is not the graph of a function because a vertical line can intersect it in two points. domain: 0, ; range: ,
24. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: ,
25. 26y x Each value of x corresponds to exactly one value of y, so this equation defines a function.
26. The equation 213
x y does not define y as a
function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 3) and (3, –3) satisfy the relation. Thus, the relation would not be a function.
27. The equation 2y x does not define y as a function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 1) and (3, −1) satisfy the relation.
28. The equation 4yx
defines y as a function
of x because for every x in the domain, which is (– , 0) (0, ) , there will be exactly one value of y.
29. In the function ( ) 4f x x , we may use
any real number for x. The domain is , .
30. 8
( )8
xf xx
x can be any real number except 8 because this will give a denominator of zero. Thus, the domain is (– , 8) (8, ).
31. 6 3f x x
In the function 6 3y x , we must have
6 3 0x . 6 3 0 6 3 2 2x x x x Thus, the domain is , 2 .
32. (a) As x is getting larger on the interval 2, , the value of y is increasing.
(b) As x is getting larger on the interval , 2 , the value of y is decreasing.
(c) f(x) is constant on (−2, 2).
In exercises 33–36, 22 3 6.f x x x
33. 23 2 3 3 3 62 9 3 3 618 9 6 15
f
34.
20.5 2 0.5 3 0.5 6
2 0.25 3 0.5 60.5 1.5 6 8
f
35. 20 2 0 3 0 6 6f
36. 22 3 6f k k k
37. 25
2 5 5 5 2 5 1x y y x y x
The graph is the line with slope 25
and
y-intercept (0, –)1. It may also be graphed using intercepts. To do this, locate the x-intercept: 0y
52
2 5 0 5 2 5x x x
38. 37
3 7 14 7 3 14 2x y y x y x
The graph is the line with slope of 37
and
y-intercept (0, 2). It may also be graphed using intercepts. To do this, locate the x-intercept by setting y = 0:
143
3 7 0 14 3 14x x x
252 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
39. 25
2 5 20 5 2 20 4x y y x y x
The graph is the line with slope of 25
and
y-intercept (0, 4). It may also be graphed using intercepts. To do this, locate the x-intercept: x-intercept: 0y
2 5 0 20 2 20 10x x x
40. 13
3y x y x
The graph is the line with slope 13
and
y-intercept (0, 0), which means that it passes through the origin. Use another point such as (6, 2) to complete the graph.
41. f(x) = x The graph is the line with slope 1 and y-intercept (0, 0), which means that it passes through the origin. Use another point such as (1, 1) to complete the graph.
42.
14
4 84 8
2
x yy xy x
The graph is the line with slope 14
and
y-intercept (0, –2). It may also be graphed using intercepts. To do this, locate the x-intercept:
0 4 0 8 8y x x
43. x = –5 The graph is the vertical line through (–5, 0).
44. f(x) = 3 The graph is the horizontal line through (0, 3).
45. 2 0 2y y The graph is the horizontal line through (0, −2).
46. The equation of the line that lies along the x-axis is y = 0.
47. Line through (0, 5), 23
m
Note that 2 23 3
.m
Begin by locating the point (0, 5). Because the
slope is 23 , a change of 3 units horizontally
(3 units to the right) produces a change of –2 units vertically (2 units down). This gives a second point, (3, 3), which can be used to complete the graph.
(continued on next page)
Chapter 2 Review Exercises 253
Copyright © 2017 Pearson Education, Inc.
(continued)
48. Line through (2, –4), 34
m
First locate the point (2, –4).
Because the slope is 34
, a change of 4 units
horizontally (4 units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (6, –1), which can be used to complete the graph.
49. through (2, –2) and (3, –4) 2 1
2 1
4 2 22
3 2 1
y ymx x
50. through (8, 7) and 12
, 2
2 11 15
2 1 2 2
2 7 9
8
2 18 69
15 15 5
y ymx x
51. through (0, –7) and (3, –7) 7 7 0
03 0 3
m
52. through (5, 6) and (5, –2)
2 1
2 1
2 6 8
5 5 0
y ymx x
The slope is undefined.
53. 11 2 3x y Solve for y to put the equation in slope-intercept form.
3112 2
2 11 3y x y x
Thus, the slope is 112
.
54. 9x – 4y = 2. Solve for y to put the equation in slope-intercept form.
9 14 2
4 9 2y x y x
Thus, the slope is 94
.
55. 2 0 2x x The graph is a vertical line, through (2, 0). The slope is undefined.
56. x – 5y = 0. Solve for y to put the equation in slope-intercept form.
15
5y x y x
Thus, the slope is 15
.
57. Initially, the car is at home. After traveling for 30 mph for 1 hr, the car is 30 mi away from home. During the second hour the car travels 20 mph until it is 50 mi away. During the third hour the car travels toward home at 30 mph until it is 20 mi away. During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home. During the last hour, the car travels 60 mi at 60 mph until it arrived home.
58. (a) This is the graph of a function because no vertical line intersects the graph in more than one point.
(b) The lowest point on the graph occurs in December, so the most jobs lost occurred in December. The highest point on the graph occurs in January, so the most jobs gained occurred in January.
(c) The number of jobs lost in December is approximately 6000. The number of jobs gained in January is approximately 2000.
(d) It shows a slight downward trend.
59. (a) We need to first find the slope of a line that passes between points (0, 30.7) and (12, 82.9)
2 1
2 1
82.9 30.7 52.24.35
12 0 12
y ymx x
Now use the point-intercept form with b = 30.7 and m = 4.35. y = 4.35x + 30.7 The slope, 4.35, indicates that the number of e-filing taxpayers increased by 4.35% each year from 2001 to 2013.
(b) For 2009, we evaluate the function for x = 8. y = 4.35(8) + 30.7 = 65.5 65.5% of the tax returns are predicted to have been filed electronically.
254 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
60. We need to find the slope of a line that passes between points (1980, 21000) and (2013, 63800)
2 1
2 1
63,800 21,000
2013 198042,800
$1297 per year33
y ymx x
The average rate of change was about $1297 per year.
61. (a) through (3, –5) with slope –2 Use the point-slope form.
1 1( )( 5) 2( 3)
5 2( 3)5 2 6
2 1
y y m x xy x
y xy x
y x
(b) Standard form: 2 1 2 1y x x y
62. (a) through (–2, 4) and (1, 3) First find the slope.
3 4 1
1 ( 2) 3m
Now use the point-slope form with 1
1 1 3( , ) (1, 3) and .x y m
1 113
1013 3
( )
3 ( 1)3( 3) 1( 1)
3 9 13 10
y y m x xy x
y xy x
y x y x
(b) Standard form: 101
3 33 10
3 10y x y xx y
63. (a) through (2, –1) parallel to 3x – y = 1 Find the slope of 3x – y = 1. 3 1 3 1 3 1x y y x y x The slope of this line is 3. Because parallel lines have the same slope, 3 is also the slope of the line whose equation is to be found. Now use the point-slope form with 1 1( , ) (2, 1) and 3.x y m
1 1( )( 1) 3( 2)
1 3 6 3 7
y y m x xy x
y x y x
(b) Standard form: 3 7 3 7 3 7y x x y x y
64. (a) x-intercept (–3, 0), y-intercept (0, 5) Two points of the line are (–3, 0) and (0, 5). First, find the slope.
5 0 5
0 3 3m
The slope is 53
and the y-intercept is
(0, 5). Write the equation in slope-
intercept form: 53
5y x
(b) Standard form: 53
5 3 5 155 3 15 5 3 15
y x y xx y x y
65. (a) through (2, –10), perpendicular to a line with an undefined slope A line with an undefined slope is a vertical line. Any line perpendicular to a vertical line is a horizontal line, with an equation of the form y = b. The line passes through (2, –10), so the equation of the line is y = –10.
(b) Standard form: y = –10
66. (a) through (0, 5), perpendicular to 8x + 5y = 3 Find the slope of 8x + 5y = 3.
8 35 5
8 5 3 5 8 3x y y xy x
The slope of this line is 85
. The slope
of any line perpendicular to this line is 58
, because 8 55 8
1.
The equation in slope-intercept form with
slope 58
and y-intercept (0, 5) is
58
5.y x
(b) Standard form: 58
5 8 5 405 8 40 5 8 40
y x y xx y x y
67. (a) through (–7, 4), perpendicular to y = 8 The line y = 8 is a horizontal line, so any line perpendicular to it will be a vertical line. Because x has the same value at all points on the line, the equation is x = –7. It is not possible to write this in slope-intercept form.
(b) Standard form: x = –7
68. (a) through (3, –5), parallel to y = 4 This will be a horizontal line through (3, –5). Because y has the same value at all points on the line, the equation is y = –5.
Chapter 2 Review Exercises 255
Copyright © 2017 Pearson Education, Inc.
(b) Standard form: y = –5
69. ( ) 3f x x
The graph is the same as that of ,y x
except that it is translated 3 units downward.
70. ( )f x x
The graph of ( )f x x is the reflection of
the graph of y x about the x-axis.
71. 2( ) 1 3f x x
The graph of 2( ) 1 3f x x is a
translation of the graph of 2y x to the left 1 unit, reflected over the x-axis and translated up 3 units.
72. ( ) 2f x x
The graph of ( ) 2f x x is the reflection
of the graph of y x about the x-axis, translated down 2 units.
73. ( ) 3f x x
To get y = 0, we need 0 3 1x 3 4.x To get y = 1, we need1 3 2x 4 5.x Follow this pattern to graph the step function.
74. 3( ) 2 1 2f x x
The graph of 3( ) 2 1 2f x x is a
translation of the graph of 3y x to the left 1 unit, stretched vertically by a factor of 2, and translated down 2 units.
75. 4 2 if 1( )
3 5 if 1x xf x x x
Draw the graph of y = –4x + 2 to the left of x = 1, including the endpoint at x = 1. Draw the graph of y = 3x – 5 to the right of x = 1, but do not include the endpoint at x = 1. Observe that the endpoints of the two pieces coincide.
256 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
76. 2 3 if 2( )
4 if 2x xf x
x x
Graph the curve 2 3y x to the left of x = 2, and graph the line y = –x + 4 to the right of x = 2. The graph has an open circle at (2, 7) and a closed circle at (2, 2).
77. if 3
( )6 if 3x xf x
x x
Draw the graph of y x to the left of x = 3,
but do not include the endpoint. Draw the graph of y = 6 – x to the right of x = 3, including the endpoint. Observe that the endpoints of the two pieces coincide.
78. Because x represents an integer, .x x
Therefore, 2 .x x x x x
79. True. The graph of an even function is symmetric with respect to the y-axis.
80. True. The graph of a nonzero function cannot be symmetric with respect to the x-axis. Such a graph would fail the vertical line test
81. False. For example, 2( )f x x is even and (2, 4) is on the graph but (2, –4) is not.
82. True. The graph of an odd function is symmetric with respect to the origin.
83. True. The constant function 0f x is both
even and odd. Because 0 ,f x f x
the function is even. Also 0 0 ,f x f x so the function is
odd.
84. False. For example, 3( )f x x is odd, and (2, 8) is on the graph but (–2, 8) is not.
85. 2 10x y
Replace x with –x to obtain 2( ) 10.x y The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to
obtain 2 2( ) 10 10.x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to
obtain 2 2( ) ( ) 10 ( ) 10.x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. The graph is symmetric with respect to the x-axis only.
86. 2 25 5 30y x Replace x with –x to obtain
2 2 2 25 5( ) 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2 2 25( ) 5 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the y-axis and x-axis, so it must also be symmetric with respect to the origin. Note that
this equation is the same as 2 2 6y x , which is a circle centered at the origin.
87. 2 3x y Replace x with –x to obtain
2 3 2 3( ) .x y x y The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace
y with –y to obtain 2 3 2 3( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y
with –y to obtain 2 3 2 3( ) ( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.
Chapter 2 Review Exercises 257
Copyright © 2017 Pearson Education, Inc.
88. 3 4y x
Replace x with –x to obtain 3 4y x . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain
3 3( ) 4 4y x y x 3 4y x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
3 3 3( ) ( ) 4 4 4.y x y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
89. 6 4x y
Replace x with –x to obtain 6( ) 4x y
6 4x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 6 ( ) 4 6 4x y x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain 6( ) ( ) 4x y 6 4x y . This equation is not equivalent to the original one, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.
90. y x
Replace x with –x to obtain ( ) .y x y x The result is not the
same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain
.y x y x The result is the same as
the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
( ) .y x y x The result is not the
same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis only.
91. y = 1 This is the graph of a horizontal line through (0, 1). It is symmetric with respect to the y-axis, but not symmetric with respect to the x-axis and the origin.
92. x y
Replace x with –x to obtain .x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
.x y x y The result is the same as
the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.
93. 2 2 0x y Replace x with −x to obtain
2 2 2 20 0.x y x y The result is
the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain
22 2 20 0.x y x y The result is
the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.
94. 22 2 4x y
Replace x with −x to obtain
2 2 222 4 2 4.x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain
22 2 4.x y The result is not the same
as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain
2 2 222 4 2 4,x y x y
which is not equivalent to the original equation. Therefore, the graph is not symmetric with respect to the origin.
95. To obtain the graph of ( )g x x , reflect the
graph of ( )f x x across the x-axis.
96. To obtain the graph of ( ) 2h x x , translate
the graph of ( )f x x down 2 units.
258 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
97. To obtain the graph of ( ) 2 4k x x ,
translate the graph of ( )f x x to the right 4
units and stretch vertically by a factor of 2.
98. If the graph of ( ) 3 4f x x is reflected about the x-axis, we obtain a graph whose equation is (3 4) 3 4.y x x
99. If the graph of ( ) 3 4f x x is reflected about the y-axis, we obtain a graph whose equation is ( ) 3( ) 4 3 4.y f x x x
100. If the graph of ( ) 3 4f x x is reflected about the origin, every point (x, y) will be replaced by the point (–x, –y). The equation for the graph will change from 3 4y x to
3( ) 4y x 3 4y x
3 4.y x
101. (a) To graph ( ) 3,y f x translate the graph of y = f(x), 3 units up.
(b) To graph ( 2),y f x translate the graph of y = f(x), 2 units to the right.
(c) To graph ( 3) 2,y f x translate the graph of y = f(x), 3 units to the left and 2 units down.
(d) To graph ( )y f x , keep the graph of
y = f(x) as it is where 0y and reflect the graph about the x-axis where y < 0.
102. No. For example suppose 2f x x and
2 .g x x Then
( )( ) ( ( )) (2 ) 2 2f g x f g x f x x
The domain and range of g are ( , ) ,
however, the domain of f is [2, ) . So,
2 2 0 1x x . Therefore, the domain of
f g is 1, . The domain of g, ( , ), is
not a subset of the domain of , 1, .f g
For Exercises 103–110, 2( ) 3 4f x x= − and 2( ) 3 4.g x x x= − −
103. 2 2
4 3 2 2
4 3 2
( )( ) ( ) ( )
(3 4)( 3 4)
3 9 12 4 12 16
3 9 16 12 16
fg x f x g xx x x
x x x x xx x x x
104. 2 2
( )(4) (4) (4)
(3 4 4) (4 3 4 4)(3 16 4) (16 3 4 4)(48 4) (16 12 4)44 0 44
f g f g
105. 2 2
( )( 4) ( 4) ( 4)
[3( 4) 4] [( 4) 3( 4) 4][3(16) 4] [16 3( 4) 4][48 4] [16 12 4]44 24 68
f g f g
106. 2 2
2 2
2 2
2
( )(2 ) (2 ) (2 )
[3(2 ) 4] [(2 ) 3(2 ) 4]
[3(4) 4] [4 3(2 ) 4]
(12 4) (4 6 4)
16 6 8
f g k f k g kk k kk k k
k k kk k
Chapter 2 Review Exercises 259
Copyright © 2017 Pearson Education, Inc.
107. 2
2
(3) 3 3 4 3 9 4(3)
(3) 9 3 3 43 3 3 427 4 23 23
9 9 4 4 4
f fg g
108.
2
2
3 1 4 3 1 4( 1)
1 3 1 41 3 1 43 4 1
undefined1 3 4 0
fg
109. The domain of (fg)(x) is the intersection of the domain of f(x) and the domain of g(x). Both have domain , , so the domain of
(fg)(x) is , .
110. 2 2
2
3 4 3 4( )
( 1)( 4)3 4
f x xxg x xx x
Because both f x and g x have domain
, , we are concerned about values of x
that make 0.g x Thus, the expression is
undefined if (x + 1)(x – 4) = 0, that is, if x = –1 or x = 4. Thus, the domain is the set of all real numbers except x = –1 and x = 4, or (– , –1) (–1, 4) (4, ).
111. 2 9f x x
( ) 2( ) 9 2 2 9f x h x h x h
( ) ( ) (2 2 9) (2 9)2 2 9 2 9 2
f x h f x x h xx h x h
Thus, ( ) ( ) 2
2.f x h f x h
h h
112. 2( ) 5 3f x x x 2
2 2( ) ( ) 5( ) 3
2 5 5 3
f x h x h x hx xh h x h
2 2 2
2 2 2
2
( ) ( )
( 2 5 5 3) ( 5 3)
2 5 5 3 5 3
2 5
f x h f xx xh h x h x x
x xh h x h x xxh h h
2( ) ( ) 2 5
(2 5)2 5
f x h f x xh h hh h
h x h x hh
For Exercises 113–118, 2( ) 2 and ( ) .f x x g x x
113. 2
( )( ) [ ( )] 2
2 2
g f x g f x g x
x x
114. 2 2( )( ) [ ( )] ( ) 2f g x f g x f x x
115. 2, so 3 3 2 1 1.f x x f
Therefore,
3 3 1g f g f g 21 1.
116. 22 , so 6 6 36.g x x g
Therefore, 6 6 36f g f g f
36 2 34 .
117. 1 1 1 2 3g f g f g g
Because 3 is not a real number, 1g f
is not defined.
118. To find the domain of ,f g we must consider the domain of g as well as the composed function, .f g Because
2 2f g x f g x x we need to
determine when 2 2 0.x Step 1: Find the values of x that satisfy
2 2 0.x 2 2 2x x
Step 2: The two numbers divide a number line into three regions.
Step 3 Choose a test value to see if it satisfies
the inequality, 2 2 0.x
Interval Test
Value Is 2 2 0x true or false?
, 2 2 22 2 0 ?2 0 True
2, 2 0 20 2 0 ?
2 0 False
2, 2 22 2 0 ?
2 0 True
The domain of f g is
, 2 2, .
260 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
119. 1 1 1 7 1 8f g f g
120. ( )(3) (3) (3) 9 9 0f g f g
121. ( )( 1) ( 1) ( 1) 3 2 6fg f g
122. (0) 5
(0) undefined(0) 0
f fg g
123. ( )( 2) [ ( 2)] (1) 2g f g f g
124. ( )(3) [ (3)] ( 2) 1f g f g f
125. ( )(2) [ (2)] (2) 1f g f g f
126. ( )(3) [ (3)] (4) 8g f g f g
127. Let x = number of yards. f(x) = 36x, where f(x) is the number of inches. g(x) = 1760x, where g(x) is the number of yards. Then
( )( ) ( ) 1760(36 )g f x g f x x 63,360 .x
There are 63,360x inches in x miles
128. Use the definition for the perimeter of a rectangle. P = length + width + length + width
( ) 2 2 6P x x x x x x This is a linear function.
129. If 343
( )V r r and if the radius is increased
by 3 inches, then the amount of volume gained is given by
3 34 43 3
( ) ( 3) ( ) ( 3) .gV r V r V r r r
130. (a) 2V r h If d is the diameter of its top, then h = d
and 2
.dr So,
32 2
2 4 4( ) ( ) ( ) .d d dV d d d
(b) 22 2S r rh
2
2 2 2
2 22 2 2
2 32 2 2
( ) 2 2 ( )d d d
d d d
S d d d
Chapter 2 Test
1. (a) The domain of 3f x x occurs
when 0.x In interval notation, this
correlates to the interval in D, 0, .
(b) The range of 3f x x is all real
numbers greater than or equal to 0. In interval notation, this correlates to the interval in D, 0, .
(c) The domain of 2 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(d) The range of 2 3f x x is all real
numbers greater than or equal to 3. In interval notation, this correlates to the interval in B, 3, .
(e) The domain of 3 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(f) The range of 3 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(g) The domain of 3f x x is all real
numbers. In interval notation, this correlates to the interval in C, , .
(h) The range of 3f x x is all real
numbers greater than or equal to 0. In interval notation, this correlates to the interval in D, 0, .
(i) The domain of 2x y is 0x because when you square any value of y, the outcome will be nonnegative. In interval notation, this correlates to the interval in D, 0, .
(j) The range of 2x y is all real numbers. In interval notation, this correlates to the interval in C, , .
2. Consider the points 2,1 and 3, 4 .
4 1 3
3 ( 2) 5m
3. We label the points A 2,1 and B 3, 4 .
2 2
2 2
( , ) [3 ( 2)] (4 1)
5 3 25 9 34
d A B
Chapter 2 Test 261
Copyright © 2017 Pearson Education, Inc.
4. The midpoint has coordinates
2 3 1 4 1 5, , .
2 2 2 2
5. Use the point-slope form with 3
1 1 5( , ) ( 2,1) and .x y m
1 13535
( )
1 [ ( 2)]
1 ( 2) 5 1 3( 2)5 5 3 6 5 3 11
3 5 11 3 5 11
y y m x xy xy x y xy x y x
x y x y
6. Solve 3x – 5y = –11 for y.
3 115 5
3 5 115 3 11
x yy xy x
Therefore, the linear function is 3 115 5
( ) .f x x
7. (a) The center is at (0, 0) and the radius is 2, so the equation of the circle is
2 2 4x y .
(b) The center is at (1, 4) and the radius is 1, so the equation of the circle is
2 2( 1) ( 4) 1x y
8. 2 2 4 10 13 0x y x y Complete the square on x and y to write the equation in standard form:
2 2
2 2
2 2
2 2
4 10 13 0
4 10 13
4 4 10 25 13 4 25
2 5 16
x y x yx x y y
x x y y
x y
The circle has center (−2, 5) and radius 4.
9. (a) This is not the graph of a function because some vertical lines intersect it in more than one point. The domain of the relation is [0, 4]. The range is [–4, 4].
(b) This is the graph of a function because no vertical line intersects the graph in more than one point. The domain of the function is (– , –1) (–1, ). The
range is (– , 0) (0, ). As x is getting
larger on the intervals , 1 and
1, , the value of y is decreasing, so
the function is decreasing on these intervals. (The function is never increasing or constant.)
10. Point A has coordinates (5, –3). (a) The equation of a vertical line through A
is x = 5.
(b) The equation of a horizontal line through A is y = –3.
11. The slope of the graph of 3 2y x is –3.
(a) A line parallel to the graph of 3 2y x has a slope of –3.
Use the point-slope form with
1 1( , ) (2,3) and 3.x y m
1 1( )3 3( 2)3 3 6 3 9
y y m x xy xy x y x
(b) A line perpendicular to the graph of
3 2y x has a slope of 13
because
13
3 1.
13
713 3
3 ( 2)
3 3 2 3 9 2
3 7
y xy x y x
y x y x
12. (a) 2, (b) 0, 2
(c) , 0 (d) ,
(e) , (f) 1,
13. To graph 2 1f x x , we translate the
graph of ,y x 2 units to the right and 1 unit
down.
262 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc.
14. ( ) 1f x x
To get y = 0, we need 0 1 1x 1 0.x To get y = 1, we need
1 1 2x 0 1.x Follow this pattern to graph the step function.
15. 12
3 if 2( )
2 if 2x
f x x x
For values of x with x < –2, we graph the horizontal line y = 3. For values of x with
2,x we graph the line with a slope of 12
and a y-intercept of (0, 2). Two points on this line are (–2, 3) and (0, 2).
16. (a) Shift f(x), 2 units vertically upward.
(b) Shift f(x), 2 units horizontally to the left.
(c) Reflect f(x), across the x-axis.
(d) Reflect f(x), across the y-axis.
(e) Stretch f(x), vertically by a factor of 2.
17. Starting with ,y x we shift it to the left 2 units and stretch it vertically by a factor of 2. The graph is then reflected over the x-axis and then shifted down 3 units.
18. 2 23 2 3x y
(a) Replace y with –y to obtain 2 2 2 23 2( ) 3 3 2 3.x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the x-axis.
(b) Replace x with –x to obtain 2 2 2 23( ) 2 3 3 2 3.x y x y
The result is the same as the original equation, so the graph is symmetric with respect to the y-axis.
(c) The graph is symmetric with respect to the x-axis and with respect to the y-axis, so it must also be symmetric with respect to the origin.
Chapter 2 Test 263
Copyright © 2017 Pearson Education, Inc.
19. 2( ) 2 3 2, ( ) 2 1f x x x g x x
(a)
2
2
2
( )( ) ( ) ( )
2 3 2 2 1
2 3 2 2 1
2 1
f g x f x g xx x x
x x xx x
(b) 2( ) 2 3 2
( )( ) 2 1
f f x x xxg g x x
(c) We must determine which values solve the equation 2 1 0.x
12
2 1 0 2 1x x x
Thus, 12
is excluded from the domain,
and the domain is 1 12 2
, , .
(d) 2( ) 2 3 2f x x x
2
2 2
2 2
2 3 2
2 2 3 3 2
2 4 2 3 3 2
f x h x h x hx xh h x h
x xh h x h
2 2
2
2 2
2
2
( ) ( )
(2 4 2 3 3 2)
(2 3 2)
2 4 2 3
3 2 2 3 2
4 2 3
f x h f xx xh h x h
x xx xh h x
h x xxh h h
2( ) ( ) 4 2 3
(4 2 3)
4 2 3
f x h f x xh h hh h
h x hh
x h
(e) 2
( )(1) (1) (1)
(2 1 3 1 2) ( 2 1 1)(2 1 3 1 2) ( 2 1 1)(2 3 2) ( 2 1)1 ( 1) 0
f g f g
(f) 2
( )(2) (2) (2)
(2 2 3 2 2) ( 2 2 1)(2 4 3 2 2) ( 2 2 1)(8 6 2) ( 4 1)4( 3) 12
fg f g
(g) 2 1 0 2 0 1g x x g
0 1 1. Therefore,
2
0 0
1 2 1 3 1 22 1 3 1 22 3 2 1
f g f gf
For exercises 20 and 21, 1f x x and
2 7.g x x
20. 2 7
(2 7) 1 2 6
f g f g x f xx x
The domain and range of g are ( , ) , while
the domain of f is [0, ) . We need to find the values of x which fit the domain of f: 2 6 0 3x x . So, the domain of f g
is [3, ) .
21. 1
2 1 7
g f g f x g x
x
The domain and range of g are ( , ) , while
the domain of f is [0, ) . We need to find the values of x which fit the domain of f:
1 0 1x x . So, the domain of g f
is [ 1, ) .
22. (a) C(x) = 3300 + 4.50x
(b) R(x) = 10.50x
(c) ( ) ( ) ( )10.50 (3300 4.50 )6.00 3300
P x R x C xx x
x
(d) ( ) 06.00 3300 0
6.00 3300550
P xx
xx
She must produce and sell 551 items before she earns a profit.
20 Copyright © 2017 Pearson Education, Inc
Chapter 2 Graphs and Functions Section 2.1 Rectangular Coordinates
and Graphs Classroom Example 1 (page 184)
(a) (transportation, $12,153)
(b) (health care, $4917)
Classroom Example 2 (page 186)
22( , ) 2 3 8 5
25 169 194
d P Q
Classroom Example 3 (page 186)
22
22
2 2
( , ) 5 0 1 2
25 9 34
( , ) 4 0 3 2
16 25 41
( , ) 4 5 3 1
81 4 85
d R S
d R T
d S T
The longest side has length 85
2 2 2?34 41 85
34 41 85
The triangle formed by the three points is not a right triangle.
Classroom Example 4 (page 187)
2 2
2 2
22
The distance between ( 2,5) and (0,3) is
2 0 5 3 4 4 8 2 2
The distance between (0,3) and (8, 5) is
8 0 5 3 64 64
128 8 2
The distance between ( 2,5) and (8, 5) is
2 8 5 5 100 100
200 10 2
P Q
Q R
P R
Because 2 2 8 2 10 2 , the points are collinear.
Classroom Example 5 (page 188)
(a) The coordinates of M are
7 ( 2) 5 13 9
, , 42 2 2
(b) Let (x, y) be the coordinates of Q. Use the midpoint formula to find the coordinates:
8 20, 4, 4
2 2
84 8 8 0
220
4 20 8 122
x y
xx x
yy y
The coordinates of Q are (0, 12).
Classroom Example 6 (page 188)
The year 2011 lies halfway between 2009 and 2013, so we must find the coordinates of the midpoint of the segment that has endpoints (2009, 124.0) and (2013, 137.4)
2009 2013 124.0 137.4,
2 2
2011, 130.7
M
The estimate of $130.7 billion is $0.1 billion more than the actual amount.
Classroom Example 7 (page 189)
Choose any real number for x, substitute the value in the equation and then solve for y. Note that additional answers are possible.
(a) x y = −2x + 5
−1 2( 1) 5 7y
0 2(0) 5 5y
3 2(3) 5 1y
Three ordered pairs that are solutions are (−1, 7), (0, 5), and (3, −1). Other answers are possible.
Section 2.2 Circles 21
Copyright © 2017 Pearson Education, Inc
(b) y 3 1x y
−9 3 39 1 8 2x
−2 3 32 1 1 1x
−1 3 31 1 0 0x
0 33 0 1 1 1x
7 3 37 1 8 2x
Ordered pairs that are solutions are (−2, −9), (−1, −2), (0, −1), (1, 0) and (2, 7). Other answers are possible.
(c) x 2 1y x
−2 2( 2) 1 3y
−1 2( 1) 1 0y
0 20 1 1y
1 21 1 0y
2 22 1 3y
Ordered pairs that are solutions are (−2, −3), (−1, 0), (0, 1), (1, 0) and (2, −3). Other answers are possible.
Classroom Example 8 (page 190)
(a) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:
52
0 2 52(0) 5 5x x
y y
Find a third point on the graph by letting x = −1 and solving for y: 2 1 5 7y .
The three points are 52
, 0 , (0, 5), and (−1, 7).
Note that (3, −1) is also on the graph.
(b) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:
33
3
0 1 1 1
0 1 0 1 1
x
y y y
Find a third point by letting x = 2 and solving
for y: 332 1 2 1 7y y y .
Find a fourth point by letting x = −2 and solving for y:
332 1 2 1 9y y y
The points to be plotted are (0, −1), (1, 0), (2, 7), and (−2, −9). Note that (−1, −2) is also on the graph.
(c) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:
2 2 2
2
0 1 1 1 1
0 1 1
x x x x
y
Find a third point by letting x = 2 and solving
for y: 22 1 3y . Find a fourth point
by letting x = −2 and solving for y:
22 1 3y
The points to be plotted are (−1, 0), (1, 0), (0, 1), (2, −3), and (−2, −3)
Section 2.2 Circles Classroom Example 1 (page 195)
(a) (h, k) = (1, −2) and r = 3
2 2 2
22 2
2 2
1 2 3
1 2 9
x h y k r
x y
x y
(b) (h, k) = (0, 0) and r = 2
2 2 2
2 2 2
2 2
0 0 2
4
x h y k r
x y
x y
22 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 2 (page 196)
(a) 2 21 2 9x y
This is a circle with center (1, −2) and radius 3.
(b) 2 2 4x y
This is a circle with center (0, 0) and radius 2.
Classroom Example 3 (page 197)
Complete the square twice, once for x and once for y:
2 2
2 2
2 2
4 8 44 0
4 4 8 16 44 4 16
2 4 64
x x y y
x x y y
x y
Because c = 64 and 64 > 0, the graph is a circle. The center is (−2, 4) and the radius is 8.
Classroom Example 4 (page 198)
2 22 2 2 6 45x y x y
Group the terms, factor out 2, and then complete the square:
2 21 92 2 3
4 4
1 945 2 2
4 4
x x y y
Factor and then divide both sides by 2:
2 2
2 2
1 32 2 50
2 2
1 325
2 2
x y
x y
Because c = 25 and 25 > 0, the graph is a circle. The
center is 1 3
,2 2
and the radius is 5.
Classroom Example 5 (page 198)
Complete the square twice, once for x and once for y:
2 2
2 2
2 2
6 2 12 0
6 9 2 1 12 9 1
3 1 2
x x y y
x x y y
x y
Because c = −2 and −2 < 0, the graph is nonexistent.
Classroom Example 6 (page 199)
Determine the equation for each circle and then substitute x = −3 and y = 4. Station A:
2 2 2
2 2 2
2 2
1 4 4
3 1 4 4 4
4 416 16
x y
Station B:
2 2 2
2 2
2 2
2 2
6 0 5
6 25
3 6 4 25
3 4 259 16 25
25 25
x y
x y
Station C:
22 2
2 2
2 2
2 2
5 2 10
5 2 100
3 5 4 2 100
8 6 10064 36 100
100 100
x y
x y
Because (−3, 4) satisfies all three equations, we can conclude that the epicenter is (−3, 4).
Section 2.3 Functions 23
Copyright © 2017 Pearson Education, Inc
Section 2.3 Functions Classroom Example 1 (page 204)
4,0 , 3,1 , 3,1M
M is a function because each distinct x value has exactly one y value.
2,3 , 3, 2 , 4,5 , 5, 4N
N is a function because each distinct x value has exactly one y value.
4,3 , 0,6 , 2,8 , 4, 3P
P is not a function because there are two y-values for x = −4.
Classroom Example 2 (page 205)
(a) Domain: {−4, −1, 1, 3} Range: {−2, 0, 2, 5} The relation is a function.
(b) Domain: {1, 2, 3} Range: {4, 5, 6, 7} The relation is not a function because 2 maps to 5 and 6.
(c) Domain: {−3, 0, 3, 5} Range: {5} The relation is a function.
Classroom Example 3 (page 206)
(a) Domain: {−2, 4}; range: {0, 3}
(b) Domain: ( , ) ; range: ( , )
(c) Domain: [−5, 5]; range: [−3, 3]
(d) Domain: ( , ) ; range: ( , 4]
Classroom Example 4 (page 207)
(a)
Not a function
(b)
Function
(c)
Not a function
(d)
Function
Classroom Example 5 (page 208)
(a) 2 5y x represents a function because y is
always found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers or ( , ) .
Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers, ( , ) .
24 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
(b) For any choice of x in the domain of 2 3y x , there is exactly one corresponding
value for y, so the equation defines a function. The function is defined for all values of x, so the domain is ( , ) . The square of any
number is always positive, so the range is [3, ) .
(c) For any choice of x in the domain of x y ,
there are two possible values for y. Thus, the equation does not define a function. The domain is [0, ) while the range is ( , ) .
(d) By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into y x corresponds to many values of y. The
ordered pairs (0, 2) (1, 1) (1, 0) (3, −1) and so on, all satisfy the inequality. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers, ( , ) .
(e) For 3
2y
x
, we see that y can be found by
dividing x + 2 into 3. This process produces one value of y for each value of x in the domain. The domain includes all real numbers except those that make the denominator equal to zero, namely x = −2. Therefore, the domain is ( , 2) ( 2, ) . Values of y can be
negative or positive, but never zero. Therefore the range is ( , 0) (0, ) .
Classroom Example 6 (page 210)
(a) 2( 3) ( 3) 6( 3) 4 13f
(b) 2( ) 6 4f r r r
(c) ( 2) 3( 2) 1 3 7g r r r
Classroom Example 7 (page 210)
(a) 2( 1) 2( 1) 9 7f
(b) ( 1) 6f
(c) ( 1) 5f
(d) ( 1) 0f
Classroom Example 8 (page 211)
(a) 2
2
2
( ) 2 3
( 5) ( 5) 2( 5) 3 12
( ) 2 3
f x x x
f
f t t t
(b) 2
2 3 6 23
2( ) 2
32 16
( 5) ( 5) 23 3
2( ) 2
3
x y y x
f x x
f
f t t
Section 2.4 Linear Functions 25
Copyright © 2017 Pearson Education, Inc
Classroom Example 9 (page 213)
The function is increasing on ( , 1), decreasing
on (–1, 1) and constant on (1, ).
Classroom Example 10 (page 213)
The example refers to the following figure.
(a) The water level is changing most rapidly from 0 to 25 hours.
(b) The water level starts to decrease after 50 hours.
(c) After 75 hours, there are 2000 gallons of water in the pool.
Section 2.4 Linear Functions Classroom Example 1 (page 220)
32
( ) 6f x x ; Use the intercepts to graph the
function. 32
(0) 0 6 6 :f y-intercept
3 32 2
0 6 6 4 :x x x x-intercept
Domain: ( , ), range: ( , )
Classroom Example 2 (page 220)
( ) 2f x is a constant function. Its graph is a
horizontal line with a y-intercept of 2.
Domain: ( , ), range: {2}
Classroom Example 3 (page 221)
x = 5 is a vertical line intersecting the x-axis at (5, 0).
Domain:{5}, range: ( , )
Classroom Example 4 (page 221)
3 4 0;x y Use the intercepts.
3 0 4 0 4 0 0 : -intercepty y y y
3 4 0 0 3 0 0 : -interceptx x x x
The graph has just one intercept. Choose an additional value, say 4, for x.
3 4 4 0 12 4 04 12 3
y yy y
Graph the line through (0, 0) and (4, −3).
Domain: ( , ), range: ( , )
26 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 5 (page 223)
(a) 4 ( 6) 10 5
2 2 4 2m
(b) 8 8 0
05 ( 3) 8
m
(c) 10 10 20
4 ( 4) 0m
the slope is
undefined.
Classroom Example 6 (page 224)
2x − 5y = 10 Solve the equation for y. 2 5 10
5 2 102
25
x yy x
y x
The slope is 2
,5
the coefficient of x.
Classroom Example 7 (page 224)
First locate the point (−2, −3). Because the slope is 43
, a change of 3 units horizontally (3 units to the
right) produces a change of 4 units vertically (4 units up). This gives a second point, (1, 1), which can be used to complete the graph.
Classroom Example 8 (page 225)
The average rate of change per year is 8658 9547 889
296.33 million2013 2010 3
The graph confirms that the line through the ordered pairs fall from left to right, and therefore has negative slope. Thus, the amount spent by the federal government on R&D for general science decreased by an average of $296.33 million (or $296,330,000) each year from 2010 to 2013.
Classroom Example 9 (page 226)
(a) ( ) 120 2400C x x
(b) ( ) 150R x x
(c) ( ) ( ) ( )150 (120 2400)30 2400
P x R x C xx x
x
(d) ( ) 0 30 2400 0 80P x x x
At least 81 items must be sold to make a profit.
Section 2.5 Equations of Lines and Linear Models
Classroom Example 1 (page 234)
( 5) 2( 3)5 2 6
2 1
y xy x
y x
Classroom Example 2 (page 234)
First find the slope: 3 ( 1) 4
4 5 9m
Now use either point for 1 1( , )x y : 49
3 [ ( 4)]9( 3) 4( 4)9 27 4 16
9 4 114 9 11
y xy xy x
y xx y
Classroom Example 3 (page 235)
Write the equation in slope-intercept form: 34
3 4 12 4 3 12 3x y y x y x
The slope is 34
, and the y-intercept is (0, −3).
Section 2.6 Graphs of Basic Functions 27
Copyright © 2017 Pearson Education, Inc
Classroom Example 4 (page 236)
First find the slope: 4 2 1
2 2 2m
Now, substitute 12
for m and the coordinates of one
of the points (say, (2, 2)) for x and y into the slope-intercept form y = mx + b, then solve for b:
12
2 2 3b b . The equation is
12
3y x .
Classroom Example 5 (page 236)
The example refers to the following figure:
(a) The line rises 5 units each time the x-value
increases by 2 units. So the slope is 52
. The
y-intercept is (0, 5), and the x-intercept is (−2, 0).
(b) An equation defining f is 52
( ) 5f x x .
Classroom Example 6 (page 238)
(a) Rewrite the equation 3 2 5x y in slope-
intercept form to find the slope: 3 52 2
3 2 5x y y x The slope is 32
.
The line parallel to the equation also has slope 32
. An equation of the line through (2, −4) that
is parallel to 3 2 5x y is 3 32 2
( 4) ( 2) 4 3y x y x
32
7y x or 3x − 2y = 14.
(b) The line perpendicular to the equation has
slope 23
. An equation of the line through
(2, −4) that is perpendicular to 3 2 5x y is 2 2 43 3 3
( 4) ( 2) 4y x y x 82
3 3y x or 2x +3y = −8.
Classroom Example 7 (page 240)
(a) First find the slope: 7703 6695
5043 1
m
Now use either point for 1 1( , )x y :
6695 504( 1)6695 504 504
504 6191
y xy x
y x
(b) The year 2015 is represented by x = 6. 504(6) 6191 9215y
According to the model, average tuition and fees for 4-year colleges in 2015 will be about $9215.
Classroom Example 8 (page 242)
Write the equation as an equivalent equation with 0 on one side.
3 2(5 ) 2 38x x x
3 2(5 ) 2 38 0x x x
Now graph the equation to find the x-intercept.
The solution set is {−4}.
Section 2.6 Graphs of Basic Functions Classroom Example 1 (page 249)
(a) The function is continuous over ( , 0) (0, )
(b) The function is continuous over its entire domain ( , ).
28 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 2 (page 252)
(a) Graph each interval of the domain separately. If x < 1, the graph of f(x) = 2x + 4 has an endpoint at (1, 6), which is not included as part of the graph. To find another point on this part of the graph, choose x = 0, so y = 4. Draw the ray starting at (1, 6) and extending through (0, 4). Graph the function for x ≥ 1, f(x) = 4 − x similarly. This part of the graph has an endpoint at (1, 3), which is included as part of the graph. Find another point, say (4, 0), and draw the ray starting at (1, 3) which extends through (4, 0).
(b) Graph each interval of the domain separately. If x ≤ 0, the graph of f(x) = −x − 2 has an endpoint at (0, −2), which is included as part of the graph. To find another point on this part of the graph, choose x = −2, so y = 0. Draw the ray starting at (0, −2) and extending through (−2, 0). Graph the function for x > 0,
2( ) 2f x x similarly. This part of the graph
has an endpoint at (0, −2), which is not included as part of the graph. Find another point, say (2, 2), and draw the curve starting at (0, −2) which extends through (2, 2). Note that the two endpoints coincide, so (0, −2) is included as part of the graph.
Classroom Example 3 (page 254)
Create a table of sample ordered pairs:
x −6 −3 32
0 32
3 6
13
2y x −4 −3 −3 −2 −2 −1 0
Classroom Example 4 (page 254)
For x in the interval (0, 2], y = 20. For x in (2, 3], y = 20 + 2 = 22. For x in (3, 4], y = 22 + 2 = 24. For x in (4, 5], y = 24 + 2 = 26. For x in (5, 6], y = 26 + 2 = 28.
Section 2.7 Graphing Techniques Classroom Example 1 (page 260)
Use this table of values for parts (a)−(c)
x 2( ) 2g x x 212
( )h x x 212
( )k x x
−2 8 2 1
−1 2 12
14
0 0 0 0
1 2 12
14
2 8 2 1
(a) 2( ) 2g x x
Section 2.7 Graphing Techniques 29
Copyright © 2017 Pearson Education, Inc
(b) 212
( )h x x
(c) 2
1( )
2k x x
Classroom Example 2 (page 262)
Use this table of values for parts (a) and (b)
x ( )g x x ( )h x x
−2 −2 2
−1 −1 1
0 0 0
1 −1 1
2 −2 2
(a) ( )g x x
(b) ( )h x x
Classroom Example 3 (page 263)
(a) x y
Replace x with –x to obtain x y . The result is not the same as the
original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain x y x y The
result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the x-axis only.
(b) 3y x
Replace x with –x to obtain 3y x
3y x . The result is the same as the
original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain 3y x . The result is not the same
as the original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to the y-axis only.
(c) 2 6x y
Replace x with –x to obtain 2( ) 6x y
2 6x y . The result is not the same as the
original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 2 ( ) 6x y
2 6x y . The result is not the same as the
original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is not symmetric with respect to either axis.
(d) 2 2 25x y
Replace x with –x to obtain 2 2( ) 25x y 2 2 25x y . The result is
the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain
2 2( ) 25x y 2 2 25x y . The result is
the same as the original equation, so the graph is symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to both axes. Note that the graph is a circle of radius 5, centered at the origin.
30 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
Classroom Example 4 (page 265)
(a) 32y x
Replace x with –x and y with –y to obtain 3 3 3( ) 2( ) 2 2y x y x y x . The
result is the same as the original equation, so the graph is symmetric with respect to the origin.
(b) 22y x
Replace x with –x and y with –y to obtain 2 2 2( ) 2( ) 2 2y x y x y x .
The result is not the same as the original equation, so the graph is not symmetric with respect to the origin.
Classroom Example 5 (page 266)
(a) 5 3( ) 2 3g x x x x
Replace x with –x to obtain 5 3
5 3
5 3
( ) ( ) 2( ) 3( )
2 3
( 2 3 ) ( )
g x x x x
x x x
x x x g x
g(x) is an odd function.
(b) 2( ) 2 3h x x
Replace x with –x to obtain 2 2( ) 2( ) 3 2 3 ( )h x x x h x h(x) is
an even function.
(c) 2( ) 6 9k x x x
Replace x with –x to obtain 2
2( ) ( ) 6( ) 9
6 9 ( ) and ( )
k x x x
x x k x k x
k(x) is neither even nor odd.
Classroom Example 6 (page 267)
Compare a table of values for 2( )g x x with 2( ) 2f x x . The graph of f(x) is the same as the
graph of g(x) translated 2 units up.
x 2( )g x x 2( ) 2f x x
−2 4 6
−1 1 3
0 0 2
1 1 3
2 4 6
Classroom Example 7 (page 268)
Compare a table of values for 2( )g x x with 2( ) ( 4)f x x . The graph of f(x) is the same as the
graph of g(x) translated 4 units left.
x 2( )g x x 2( ) ( 4)f x x
−7 49 9
−6 36 4
−5 25 1
−4 16 0
−3 9 1
−2 4 4
−1 1 9
Classroom Example 8 (page 269)
(a) 2( ) ( 1) 4f x x
This is the graph of 2( )g x x , translated one
unit to the right, reflected across the x-axis, and then translated four units up.
x 2( )g x x 2( ) ( 1) 4f x x
−2 4 −5
−1 1 0
0 0 3
1 1 4
2 4 3
3 9 0
4 16 −5
(continued on next page)
Section 2.8 Function Operations and Composition 31
Copyright © 2017 Pearson Education, Inc
(continued)
(b) ( ) 2 3f x x
This is the graph of ( )g x x , translated three
units to the left, reflected across the x-axis, and then stretched vertically by a factor of two.
x ( )g x x ( ) 2 3f x x
−6 6 −6
−5 5 −4
−4 4 −2
−3 3 0
−2 2 −2
−1 1 −4
0 0 −6
(c) 12
( ) 2 3h x x
This is the graph of ( )g x x , translated two
units to the left, shrunk vertically by a factor of 2, and then translated 3 units down.
x ( )g x x 12
( ) 2 3h x x
−2 undefined −3
−1 undefined −2.5
0 0 12
2 3 2.3
2 2 1.4 −2
6 6 2.4 12
8 3 1.6
7 7 2.6 −1.5
Classroom Example 9 (page 270)
The graphs in the exercises are based on the following graph.
(a) ( ) ( ) 2g x f x
This is the graph of f(x) translated two units down.
(b) ( ) ( 2)h x f x
This is the graph of f(x) translated two units right.
32 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
(c) ( ) ( 1) 2k x f x
This is the graph of f(x) translated one unit left, and then translated two units up.
(d) ( ) ( )F x f x
This is the graph of f(x) reflected across the y-axis.
Section 2.8 Function Operations and Composition
Classroom Example 1 (page 278)
For parts (a)−(d), ( ) 3 4f x x and 2( ) 2 1g x x
(a) (0) 3(0) 4 4f and 2(0) 2(0) 1 1g , so
( )(0) 4 1 5f g
(b) (4) 3(4) 4 8f and 2(4) 2(4) 1 31g ,
so ( )(4) 8 31 23f g
(c) ( 2) 3( 2) 4 10f and 2( 2) 2( 2) 1 7g , so
( )( 2) ( 10)(7) 70fg
(d) (3) 3(3) 4 5f and 2(3) 2(3) 1 17g ,
so 5
(3)17
f
g
Classroom Example 2 (page 279)
For parts (a)−(e), 2( ) 3f x x x and ( ) 4 5g x x
(a) 2 2( )( ) ( 3 ) (4 5) 5f g x x x x x x
(b) 2 2( )( ) ( 3 ) (4 5) 7 5f g x x x x x x
(c) 2
3 2 2
3 2
( )( ) ( 3 )(4 5)
4 5 12 15
4 7 15
fg x x x x
x x x x
x x x
(d) 2 3
( )4 5
f x xx
g x
(e) The domains of f and g are both ( , ) . So,
the domains of f + g, f − g, and fg are the intersection of the domains of f and g,
( , ) . The domain of fg
includes those
real numbers in the intersection of the domains of f and g for which ( ) 4 5 0g x x
54
x . So the domain of fg
is
5 54 4
, , .
Classroom Example 3 (page 280)
(a)
From the figure, we have f(1) = 3 and
g(1) = 1, so (f + g)(1) = 3 + 1 = 4. f(0) = 4 and g(0) = 0, so (f − g)(0) = 4 − 0 = 4. f(−1) = 3 and g(−1) = 1, so (f g)(−1) = (3)(1) = 3
f(−2) = 0 and g(−2) = 2, so 02
( 2) 0fg
.
(b) x f(x) g(x)
−2 −5 0
−1 −3 2
0 −1 4
1 1 6
From the table, we have f(1) = 1 and g(1) = 6, so (f + g)(1) = 1 + 6 = 7. f(0) = −1 and g(0) = 4, so (f − g)(1) = −1 − 4 = −5. f(−1) = −3 and g(−1) = 2, so (f g)(−1) = (−3)(2) = −6 f(−2) = −5 and g(−2) = 0, so
50
( 2)fg
fg
is undefined.
Section 2.8 Function Operations and Composition 33
Copyright © 2017 Pearson Education, Inc
(c) ( ) 3 4, ( )f x x g x x
From the formulas, we have (1) 3(1) 4 7f and (1) 1 1g , so
(f + g)(1) = 7 + (−1) = 6. (0) 3(0) 4 4f and (0) 0 0g , so
(f − g)(1) = 4 − 0 = 4. ( 1) 3( 1) 4 1f and ( 1) 1 1g ,
so (fg)(−1) =(1)(−1) = −1. ( 2) 3( 2) 4 2f and
( 2) 2 2g , so 2
( 2) 1.2
f
g
Classroom Example 4 (page 281)
Step 1: Find ( )f x h : 2
2 2
2 2
( ) 3( ) 2( ) 4
3( 2 ) 2 2 4
3 6 3 2 2 4
f x h x h x h
x xh h x h
x xh h x h
Step 2: Find ( ) ( )f x h f x :
2 2 2
2
( ) ( )
(3 6 3 2 2 4) (3 2 4)
6 3 2
f x h f x
x xh h x h x x
xh h h
Step 3: Find the difference quotient: 2( ) ( ) 6 3 2
6 3 2f x h f x xh h h
x hh h
Classroom Example 5 (page 283)
For parts (a) and (b), ( ) 4f x x and 2
( )g xx
(a) First find g(2): 2
(2) 12
g . Now find
( )(2) ( (2)) (1) 1 4 5f g f g f
(b) First find f(5): (5) 5 4 9 3f . Now
find 23
( )(5) ( (5)) (3)g f g f g
The screens show how a graphing calculator
evaluates the expressions in this classroom example.
Classroom Example 6 (page 283)
For parts (a) and (b), ( ) 1f x x and
( ) 2 5g x x
(a) ( )( ) ( ( )) (2 5) 1 2 4f g x f g x x x
The domain and range of g are both ( , ) .
However, the domain of f is [1, ) . Therefore,
( )g x must be greater than or equal to 1:
2 5 1 2x x . So, the domain of f g
is [ 2, ) .
(b) ( )( ) ( ( )) 2 1 5g f x g f x x
The domain of f is [1, ) , while the range of f
is [0, ) . The domain of g is ( , ) .
Therefore, the domain of ( )g f is restricted
to that portion of the domain of g that intersects with the domain of f, that is [1, ) .
Classroom Example 7 (page 284)
For parts (a) and (b), 5
( )4
f xx
and 2
( )g xx
(a) 5 5
( )( ) ( ( ))2 4 2 4
xf g x f g x
x x
The domain and range of g are both all real numbers except 0. The domain of f is all real numbers except −4. Therefore, the expression for ( )g x cannot equal −4. So,
2 14
2x
x . So, the domain of f g
is the set of all real numbers except for 12
,
and 0. This is written
1 12 2
, , 0 0,
34 Chapter 2 Graphs and Functions
Copyright © 2017 Pearson Education, Inc
(b) 2 2 8
( )( ) ( ( ))5 ( 4) 5
xg f x g f x
x
The domain of f is all real numbers except −4, while the range of f is all real numbers except 0. The domain and range of g are both all real numbers except 0, which is not in the range of f. So, the domain of g f is the set of all real
numbers except for −4. This is written , 4 4,
Classroom Example 8 (page 285)
( ) 2 5f x x and 2( ) 3g x x x 2
2
2
2 2
2
( )( ) (2 5) 3(2 5) (2 5)
3(4 20 25) 2 5
12 58 70
( )( ) (3 ) 2(3 ) 5
6 2 5
g f x g x x x
x x x
x x
f g x f x x x x
x x
In general, 2 212 58 70 6 2 5,x x x x so
( )( ) ( )( )g f x f g x .
Classroom Example 9 (page 286)
2( )( ) 4(3 2) 5(3 2) 8f g x x x
Answers may vary. Sample answer: 2( ) 4 5 8f x x x and ( ) 3 2g x x .
College Algebra and Trigonometry 6th Edition Lial Solutions ManualFull Download: http://testbanklive.com/download/college-algebra-and-trigonometry-6th-edition-lial-solutions-manual/
Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com